### Relation between field and potential - class-XI

 Description: relation between field and potential Number of Questions: 89 Created by: Tanuja Atwal Tags: coulomb's law electricity and magnetism physics electrostatics
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If we move in a direction opposite to the electric lines of force:

1. electrical potential decreases

2. electrical potential increases

3. electrical potential remains uncharged

4. nothing can be said.

Correct Option: B

A uniform wire $10 \,cm$ long is carrying a steady current. The potential drop across it is $10V$. The electric field inside it is _____

1. zero

2. $1Nm^{-1}$

3. $10 \,Vm^{-1}$

4. $100 \,Vm^{-1}$

Correct Option: D
Explanation:

$E = \dfrac{V}{r} = \dfrac{10V}{10 \times 10^{-2}m} = \dfrac{10 \times 100}{10} = 100 \dfrac{v}{m}$

The electric potential while moving along the lines of force

1. decreases

2. increases

3. remains constant

4. becomes infinite

Correct Option: A

$E=-\dfrac{dV}{dr}$, here negative sign signified that

1. E is opposite to V

2. E is negative

3. E increases when V decreases

4. E is directed in the direction of decreasing V

Correct Option: D
Explanation:

The negative sign is just a convention and it signifies that the direction of E is opposite to the direction in which potential increases.

The ratio of electric force $( F _e )$ to gravitational force acting between two electrons will be:

1. $1 \times 10^{36}$

2. $2 \times 10^{39}$

3. $2.5\times 10^{39}$

4. $3 \times 10^{39}$

Correct Option: C

Electric potential at ( x, y, z ) is given as $V$= $- x ^ { 2 } y \sqrt { z }$ Find the electrical field at (2 ,1, 1)

1. $4 \hat { i } + 4 \hat { j } + 4 \hat { k }$

2. $- 4 \hat { i } - 4 \hat { j } - 2 \hat { k }$

3. $- 4 \hat { i } - 4 \hat { j } - 4 \hat { k }$

4. $4 \hat { 1 } + 4 \hat { j } + 2 \hat { k }$

Correct Option: D

The electric field and the electric potential at a point inside a shell are E and V respectively. Which of the following is correct?

1. If $E=0$, V must be zero.

2. If $V=0$, E must be zero.

3. If $E\neq 0$, V cannot be zero.

4. None of these

Correct Option: D
Explanation:

In a shell $E=0$ but $V \neq 0$.
Along the equatorial line of a dipole, $V=0$ but $E \neq 0$.

A charge of $6.25\mu C$ in an electric field is acted upon by a force $2.5N$. The potential gradient at this point is

1. $4\times 10^{5}V / m$

2. $4\times 10^{6}V / m$

3. $2.5\times 10^{-6}V / m$

4. $4\times 10^{7}V / m$

Correct Option: A
Explanation:

given force$= 2.5$ $\mu$
we know $F = Eq$
$\Rightarrow 2.5=E( 6.25\ \mu c)$

$\Rightarrow E=\dfrac{2.5}{6.25\times 10^{-6}}$

$\Rightarrow E=4\times 10^5\ V/m$
$\therefore\ Potential\ gradient\ = \dfrac{dV}{dx}=E = 4\times 10^5\ V/m$

Electric potential $V$ at some point in space is zero. Then at that point :

1. Electric intensity is necessarily zero.

2. Electric intensity is necessarily non zero.

3. Electric intensity may or may not be zero.

4. Electric intensity is necessarily infinite.

Correct Option: C
Explanation:
$E=-\dfrac { dV }{ dl }$
where $E=$ Electric field intensity ;  $V=$ Electric Potential ; $l=$ distance traveled in direction of electric field
Electric intensity at a point is the negative of rate of change of the electric potential at a point. So if a function is zero at a given point, the slope will not necessarily be zero.

In a uniform electric field, the potential is $10V$ at the origin of coordinates, and $8V$ at each of the points $(1,0,0),(0,1,0)$ and $(0,0,1)$. The potential at the point $(1,1,1)$ will be:

1. $0$

2. $4V$

3. $8V$

4. $10V$

Correct Option: B
Explanation:
The field is uniform. Hence ðv/ðr=constant=p(let)Hence v=V°+p(i+j+k)Now at origin v=V°=10voltAlso at the 3 points the value of the voltage are 8volt each. Hence p= -2At (1,1,1) The voltage is=10 -2(1+1+1)                                 =4 volt(ans)

An electric field is represented by $E$, where $A=10\ V/{m}^{2}$. The electric potential at the origin with respect to the point $(10,20)m$ will be $V$ $(0,0)=.......\ volt$.

1. $200$

2. $300$

3. $400$

4. $500$

Correct Option: D

A force of 3000 N is acting on a charge of 4 coloumb moving in a uniform electric field. The potential difference between two point at a distance of 1 cm in this field is

1. 10 V

2. 90 V

3. 750 V

4. 9000 V

Correct Option: C

Electric potential is given by $V=6x-8{xy}^{2}$. Then electric force acting on $2\ C$ point charge placed at the origin will be

1. $2\ N$

2. $6\ N$

3. $8\ N$

4. $12\ N$

Correct Option: D
Explanation:

$V=6x-8xy^2$

$Q=2C$
$\begin{array}{l} { E _{ x } }=-\dfrac { { dv } }{ { dx } } =6-8 x { y^{ 2 } }=6 \ { E _{ y } }=-\dfrac { { dv } }{ { dy } } =-8\times x\times 2y=0 \ E=\sqrt { { E _{ x } }^{ 2 }+{ E _{ y } }^{ 2 } } =\sqrt { { { \left( 6 \right) }^{ 2 } }+0 } =6 \ F=QE \ F=2\times 6=12N \end{array}$

The electrostatic potential $V$ at any point (x, y, z) in space is given by $V = 4x^2$

1. The y-and z-components of the electrostatic field at any point are zero.

2. The x-component of electric field an any point is given by $(-8x \hat{i})$

3. The x-component  of electric field at $(1, 0, 2)$ is $(-8\hat{i})$

4. The y-and z-components of the field are constant in magnitude.

Correct Option: A,B,C,D
Explanation:
We have $V = 4x^2$
So, the $x , y$ and $z$ components of the electrostatic field are

$E _x = \dfrac{-\partial V}{\partial x} = -8x$

$E _y = \dfrac{-\partial V}{\partial y} = 0$

$E _z = \dfrac{-\partial V}{\partial z} = 0$

So, $\overrightarrow{E} = E _x\hat{i} + E _y \hat{j} + E _z\hat{k} = -8x\hat{i}$.
The electrostatic field at $(1, 0, 2)$ is $\overrightarrow{E} = (-8)\hat{i} \,V/m$.

Two conducting shells of radii $2\ cm$ and $3\ cm$ are separately charged by $10\ V$ and $5\ V$ potential, respectively. Now smaller shell is placed inside bigger shell, and  then connected by a wire. What will be potential at the surface of smaller shell ?

1. zero

2. $\dfrac{35}{3}\ volt$

3. $\dfrac{25}{3}\ volt$

4. $\dfrac{10}{3}\ volt$

Correct Option: C

If on the x-axis electric potential decreases uniformly from 60 V to 20 V between x = -2 m to x = +2 m, then the magnitude of electric field at the origin

1. Must be 10 V/m

2. May be greater than 10 V/m

3. Is zero

4. Is 5 V/m

Correct Option: A

$64$ charged drops coalesce to form a bigger charged drop. The potential of bigger drop will be times that of smaller drop-

1. $4$

2. $16$

3. $64$

4. $8$

Correct Option: B

A uniform electric field $10N/C$ exists in the vertically downward direction, the increase in the electric potential as one goes through a height of $50cm$ is:

1. $20J$

2. $\dfrac{1}{5}J$

3. $5J$

4. $\dfrac{1}{20}J$

Correct Option: A
Explanation:
Electric field $=10N/C$
Vertically downward direction electric potential as one goes through $h=50cm$ $=50\times { 10 }^{ -2 }m$
Now, $V=E/d$
$=10/50\times { 10 }^{ -2 }=\dfrac { 100 }{ 5 } =20J$

In an electric field the potential at a point is given by the following relation $V = \dfrac{343}{r}$ where r is distance from the origin. The electric field at $r = 3\hat i + 2\hat j + 6\hat k$ is:

1. $21\hat i + 14\hat j + 42\hat k$

2. $3\hat i + 2\hat j + 6\hat k$

3. $\dfrac{1}{7}(3\hat i + 2\hat j + 6\hat k )$

4. $-(3\hat i + 2\hat j + 6\hat k )$

Correct Option: D
Explanation:

B. $3i+2j+6k$

Formula,

$E=\dfrac{V}{|\vec{r}|}\cdot \hat{r}$

$E=\dfrac{343}{|\vec{r}|^2}\cdot \dfrac{3i+2j+6k}{|r|}$

$=\dfrac{343}{7^2}\cdot \dfrac{3i+2j+6k}{7}$

$=3i+2j+6k$

The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero, the electric potential at a distance r?

1. Is uniform in the region

2. Is proportional to r

3. Is proportional to $r^2$

4. Increases as one goes away from the origin

Correct Option: C
Explanation:

$\quad E∝r\quad and\quad V=0\quad at\quad r=0$

$E=kr$
$E=\frac { -dv }{ dr }$
$V=-int{Edr}$
$V=-int { Krdr}$
$V=-k\frac { { r }^{ 2 } }{ 2 } +C$
$V=-k\frac { { r }^{ 2 } }{ 2 }$
$V=0\quad r=0\quad C=0$
$V=0\quad r=0\quad C=0$
v is proportional to ${ r }^{ 2 }$

In a certain region of space, the potential is given by $V=k\left[ { 2x }^{ 2 }-{ y }^{ 2 }+{ z }^{ 2 } \right]$. The electric field at the point$(1,1,1)$ has magnitude :

1. $k\sqrt { 6 }$

2. $2k\sqrt { 6 }$

3. $2k\sqrt { 3 }$

4. $4k\sqrt { 3 }$

Correct Option: B
Explanation:

Given, $V=k[2x^2-y^2+z^2]$

Electric field , $\vec{E}=-(\dfrac{dV}{dx}\hat i+\dfrac{dV}{dy}\hat j+\dfrac{dV}{dz}\hat {k})$

or,$\vec{E}=-k(4x\hat i-2y\hat j+2z\hat k)$

or,$\vec{E} _{(1,1,1)}=-k(4\hat i-2\hat j+2\hat k)$

Magnitude of electric field$=|\vec{E} _{(1,1,1)}|=\sqrt{k^2(16+4+4)}=k\sqrt {24}=2k\sqrt 6$

Let V be electric potential and E the magnitude of the electric field. At a given position, which of the statement is true

1. E is always zero where V is zero

2. V is always zero where E is zero

3. E can b zero where V is non zero

4. E is always nonzero where V is nonzero

Correct Option: A

Two plates are 1 cm apart and the potential difference between them is 10 volt. The electric field between the plates is

1. 10 N/C

2. 250 N/C

3. 500 N/C

4. 1000 N/C

Correct Option: D

The equation of an equipotential line in an electric field is $y=2x$, then the electric field strength vector at $(1,2)$ may be :

1. $4\hat { i } +3\hat { j }$

2. $4\hat { i } +8\hat { j }$

3. $8\hat { i } +4\hat { j }$

4. $-8\hat { i } +4\hat { j }$

Correct Option: D
Explanation:

Now equation of equipotential surface is $y=2x$
Now electric field along the euipotential surface should be zero
therefore angle made by equipotential surface with x-axis is $tan^{-1} { (2) }$
Now since net electric field should be perpendicular to the equipotential surface
therefore for any electric field which makes an angle $tan^{-1} { (-1/2) }$ with x-axis can be the electric field at point $(1,2)$ which is true only for option (D)

because for two perpendicular line, product of their slope should be equal to -1 i.e., $m _1 \times m _2=-1$

Two plates are at potentials $-10 V$ and $+30 V$. If the separation between the plates is $2 cm$ then the electric field between them will be

1. 2000 V/m

2. 1000 V/m

3. 500 V/m

4. 3000 V/m

Correct Option: A
Explanation:

Given,

$d=2cm$
$V _2-V _1=+30V-(-10)=40V$
Electric field, $E=\dfrac{V _2-V _1}{d}$
$E=\dfrac{40}{2\times 10^{-2}}=2000V/m$
The correct option is A.

In a certain region the electric potential at a point $(x, y, z)$ is given by the potential function $V = 2x + 3y - z$. Then the electric field in this region will :

1. increase with increase in x and y

2. increase with increase in y and z

3. increase with increase in z and x

4. remain constant

Correct Option: D
Explanation:

$V=2x+3y-z$

$E _x=-\dfrac{dV}{dx}=-2, E _y=-\dfrac{dV}{dy}=-3$ and $E _z=-\dfrac{dV}{dz}=1$

As the field components are independent of x,y and z so the field remains constant.

Which of the following is true for uniform electric field ?

1. all points are at the same potential

2. no two points can have the same potential

3. pairs of points separated by the different distance must have the same difference in potential

4. none of the above

Correct Option: D
Explanation:

uniform electric field means the electric field vector does not vary with positive and electric field lines are parallel and equally spaced. the statement given define the equipotential surface, so answer is (d) -none of these.

The electric field and the electric potential at a point are E and V respectively. Then, the incorrect statements are :

1. If E $=$ 0, V must be zero.

2. If V $=$ 0, E must be zero.

3. If E $\neq$ 0, V cannot be zero.

4. If V $\neq$ 0, E cannot be zero.

Correct Option: A,B,C,D
Explanation:

The electric field is $E=-\dfrac{dV}{dx}$
If $V=0$, we can not say $E$ must be zero, we say only $E$ may be zero.
If $V \neq 0$, $E$ must be zero when $V$ is max i.e, $\dfrac{dV}{dx}=0$ For example, inside the conductor $E=0,$ but $V \neq 0$
If $E \neq 0$ , $V$  may be zero when two equal and opposite charges separated by a distance and  at the midpoint in between the charges field is non-zero but potential is zero.

A charge of $6.76$ $\mu$C in an electric field is acted upon by a force of $2.5 N$. The potential gradient at this point is :

1. $3.71 \times 10^{15} Vm^{-1}$

2. $-3.71 \times 10^{12} Vm^{-1}$

3. $3.71 \times 10^{10} Vm^{-1}$

4. $-3.71 \times 10^{5} Vm^{-1}$

Correct Option: D
Explanation:

$F=qE \Rightarrow E=\dfrac{F}{q}=\dfrac{2.5}{6.76\times 10^{-6}}=3.7\times 10^5$

The potential gradient is the electric field i.e, $E=-\nabla V=-3.7 \times 10^5$

The electric potential decreases uniformly from $120$ V to $80$ V as one moves on the X-axis from x $=$ -1 cm to x $=$ $+1$ cm. The electric field at the origin :

1. must be equal to $20$ V/cm

2. may be equal to $20$ V/cm

3. may be greater than $20$ V/m

4. may be less than $20$ V/cm

Correct Option: A
Explanation:

As the electric potential decreases uniformly so the electric field is uniform along the x axis.

$E=-\dfrac{dV}{dx}=\dfrac{120-80}{1-(-1)}=20$V/m. It must be equal to 20 V/m at the origin.

The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero,

1. it is uniform in the region

2. it is proportional to r

3. it is proportional to r$^2$

4. it increases as one goes away from the origion

Correct Option: C
Explanation:

Electric field is directly propotional to $r$, therefore
$E=kr$
We know that
$\ V=-\int _{ 0 }^{ r }{ \overrightarrow { E. } \overrightarrow { dr } }$
which gives
$V=-\dfrac { k{ r }^{ 2 } }{ 2 }$ since V at $r=0$ is $0$
Hence V is proportional to ${ r }^{ 2 }$

A uniform electric field of $20$ NC$^{-1}$ exists along the x-axis in space. The potential difference V$_B-$V$_A$ for the point A $=$ $(4 m, 2m)$ and B $=$ $(6m, 5m)$ is:

1. $20$ $\sqrt{13}$ V

2. $- 40 V$

3. zero V

4. none of the above

Correct Option: B
Explanation:

Here, $\vec{E}=20 \hat i \Rightarrow E _x=20$
Potential difference ,$V _B-V _A=-\int _A^B Edr=-\int _4^6E _x dx=-20\int _4^6 dx=-20(6-4)=-40 V$

The electric field at the origin is along the positive X-axis. A small circle is drawn with the centre at the origin cutting the axes at points A, B, C and D having coordinates $(a, 0), (0, a), (-a, 0), (0, -a)$ respectively. Out of the given points on the periphery of the circle, the potential is minimum at :

1. A

2. B

3. C

4. D

Correct Option: A
Explanation:

The relation between electric field and potential is given by $\vec{E} = \displaystyle -\frac{\partial V}{\partial x}\hat{i} -\frac{\partial V}{\partial y}\hat{j}$

Given that, at origin, electric field is along positive x-axis.
Thus, $\displaystyle \frac{\partial V}{\partial x} < 0$ and $\displaystyle \frac{\partial V}{\partial y} = 0$
Thus, $V$ decreases in the positive x-direction and remains constant in y-direction.
Hence, minimum $V$ occurs at $(a,0)$ i.e., $A$

It is found that air breaks down electrically, when the electric field is $3 \times 10^{6} \mathrm{V} / \mathrm{m} .$ What is the potential to which a sphere of radius $1 \mathrm{m}$ can be raised, before sparking takes place?

1. $V=10^{6} \mathrm{V}$

2. $V=2 \times 10^{6} \mathrm{V}$

3. $V=3 \times 10^{6} \mathrm{V}$

4. $V=4 \times 10^{6} \mathrm{V}$

Correct Option: C

In moving from A to B along an electric field line, the wok done by the electric field on an electron is $6.4 \times 10^{-19}$ J. If $\phi _1$ and $\phi _2$ are equipotential surfaces, then the potential difference $V _b-V _A$ is

1. -4V

2. 4V

3. zero

4. 6.4 V

Correct Option: B

The equation of an equipotential line is an electric field is y = 2x, then the electric field strength vector at (1, 2) may be

1. $4\vec{i} + 3\vec{j}$

2. $4\vec{i} + 8\vec{j}$

3. $8\vec{i} + 4\vec{j}$

4. $-8\vec{i} + 4\vec{j}$

Correct Option: B

The electric potential in a certain region along the x-axis varies with x according to the relation $V(x) = 5 - 4x^2$. Then, the correct statement is :

1. the potential difference between the points $x =1$m and $x=2$m is $12$ Volt

2. the force experienced by a Coulomb of charge placed at $x =1$ m is $8$ Newton

3. the electric field components along Y and Z direction are zero

4. all of the above

Correct Option: D
Explanation:

$V(x)=5-4x^2$

$V(1)=5-4=1 V, V(2)=5-4(2^2)=-11 V$

Potential difference between $x=1 m$ and $x=2 m$ is $V _{12}=V _1-V _2=1-(-11)=12 V$

here, $E _x=-\dfrac{dV}{dx}=8x, E _y=-\dfrac{dV}{dy}=0$ and $E _z=-\dfrac{dV}{dz}=0$

The electric force on $1$ coulomb charge at $x=1$ is $F=qE _x=1(8)=8 N$

A point charge q moves from point P to a point S along a path PQRS in a uniform electric field E pointing parallel to the x-axis. The coordinates of P, Q. R and S are $(a, b, 0), (2a, 0, 0), (a, -b, 0)$ and $(0, 0, 0)$. The work done by the field in the above process is :

1. $zero$

2. $qEB$

3. $qEa$

4. $-qEa$

Correct Option: D
Explanation:

As the field E is uniform, so E is constant at every point.
As E is directed parallel to x axis, so $\vec{E}=E\hat i$
The work done , $W=\int \vec{F}.\vec{dr}=\int qE\hat i.(\hat{i}dx+\hat{j}dy+\hat{k}dz)$
$W=qE\int dx=qE[\int _a^{2a}dx+\int^a _{2a}dx+\int _a^{0}dx]=qE[2a-a+a-2a+0-a]=-qEa$

In a certain region of space, the potential is given by : $V = k {[2x^2 - y^2 + z^2]}$. The electric field at the point (1, 1, 1) has magnitude =

1. $k\sqrt{6}$

2. $2k\sqrt{6}$

3. $2k\sqrt{3}$

4. $4k\sqrt{3}$

Correct Option: B

A charge of 3C moving in a uniform electric field experiences a force of $3000 N$. The potential difference between two points situated in the field at a distance $1 cm$ from each other will be

1. $10 V$

2. $90 V$

3. $1000 V$

4. $9000 V$

Correct Option: A

The potential at a point $x$ (measured in $\mu m )$ due to somecharges situated on the $x$ -axis is given by $V ( x ) = 20 / \left( x ^ { 2 } - 4 \right)$Volts. The electric field $E$ at $x = 4 \mu m$ is given by

1. 5$/ 3$ Volt / \mum and in the -ve $x$ direction

2. 5$/ 3$ Volt $/ \mu m$ and in the +ve $x$ direction

3. 10$/ 9$ Volt / \mum and in the -ve $x$ direction

4. 10$/ 9$ Volt $/ \mu m$ and in the +ve $x$ direction

Correct Option: B

Variation in potential is maximum if one goes :

1. along the line of force

2. perpendicular to the line of force

3. in any direction

4. none of these

Correct Option: A
Explanation:

$dV=-\vec{E}.d\vec{r}=-Edr cos\theta$

Hence, variation will be maximimum for $\theta=0^{o}$ or $180^{o}$, that is variation $dV$ is maximum along line of field or say line of force.

The electric field lines are closer together near object $A$ than they are near object $B$. We can conclude that :

1. the potential near $A$ is greater than the potential near $B$

2. the potential near $A$ is less than the potential near $B$

3. the potential near $A$ is equal to the potential near $B$

4. nothing about the relative potentials near $A$ and $B$

Correct Option: D
Explanation:

Potential decreases in the direction of electric field. So it depends  on whether the lines of forces are from $A$ to $B$ or from $B$ to $A$.

There is an electric field $E$ in the x-direction. If the work done by the electric field in moving a charge of $0.2 C$ through a distance of $2 m$ along a line making an angle $60^{\circ}$ with the x-axis is $4 J$, then what is the value of $E$?

1. $\displaystyle \sqrt3 NC^{-1}$

2. $\displaystyle 4 NC^{-1}$

3. $\displaystyle 5 NC^{-1}$

4. $\displaystyle 20 NC^{-1}$

Correct Option: D
Explanation:

$\displaystyle F = qE$
work will only be done in moving the charged particle in $x$ direction

work done in moving the charge in y-direction will be $0$
Work done , $W=\int \vec{F}.\vec{dr}$

$\displaystyle W = qE \times 2 cos 60^{\circ}$

or $\displaystyle 4 = 0.2E\times 2 \times \dfrac{1}{2}$

$\implies E = 20 NC^{-1}$

Charge $Q$ is given a displacement $\displaystyle \vec{r} = a\hat{i}+b\hat{j}$ in an electric field $\displaystyle \vec{E} = E _1\hat{i}+E _2\hat{j}$. The work done is :

1. $\displaystyle Q(E _1a+E _2b)$

2. $\displaystyle Q\sqrt{(E _1a)^2+(E _2b)^2}$

3. $\displaystyle Q (E _1+E _2) \sqrt{a^2+b^2}$

4. $\displaystyle Q \sqrt{(E _1^2+E^2 _2)^2} \sqrt{a^2+b^2}$

Correct Option: A
Explanation:

Work done in the presence of electric field E is $W=\vec F. \vec r = q\vec E.\vec r$
$W=Q[(E _1\hat{i}+E _2\hat{j}).( a\hat{i}+b\hat{j})]$
$W=Q(E _1a+E _2b)$

The electric potential decreases uniformly from $120V$ to $80V$ as one moves on the x-axis from $x=-1cm$ to $x=+1cm$. The electric field at the origin

1. must be equal to $20V{cm}^{-1}$

2. may be equal to $20V{cm}^{-1}$

3. may be greater than $20V{cm}^{-1}$

4. may be less than $20V{cm}^{-1}$

Correct Option: B,C

The electric potential decreases uniformly from 120 V to 80 V as one moves on the x-axis from $x = -1\ cm$ to $x = +1 \ cm$. The electric field at the origin.

1. must be equal to 20 Vcm$^{-1}$

2. must be equal to 20 Vm$^{-1}$

3. greater than or equal to 20 Vcm$^{-1}$

4. may be less than 20 Vcm$^{-1}$

Correct Option: A,C
Explanation:

$dv=-\vec E.\vec dx=-E dx \cos\theta$

$\displaystyle E = -\dfrac{dV}{\cos\theta dx}$

$\cos\theta\approx 1$, if we take $\cos\theta =1$,then

$E _{min}= -\dfrac{80-120}{1-(-1)} =\dfrac{40}{2}= 20 V cm^{-1}$

Hence E would be greater than or equal to $20$

Mark the correct statement:

1. If $E$ is zero at a certain point, then $V$ should be zero at that point

2. If $E$ is not zero at a certain point, then $V$ should not be zero at that point

3. If $V$ is zero at a certain point, then $E$ should be zero at that point

4. If $V$ is zero at a certain point, then $E$ may or maynot be zero

Correct Option: D
Explanation:

Since $E=\dfrac{-dV}{dr}$

A zero potential at a point never means that the electric field is also zero at a point always.
$E=0$ when $V=$ constant.
If $V=0$; then $E$ may or may not be zero.
Similarly, if $E=0$; then $V$ must either be a constant or may be zero.
so correct option is(d).

For a uniform electric field $\vec{E}=E _{0}(\hat{i})$, if the electric potential at x=0 is zero, then the value of electric potential at x=+x will be .......

1. $xE _{0}$

2. -$xE _{0}$

3. $x^{2}E _{0}$

4. -$x^{2}E _{0}$

Correct Option: B

The potential $V$ is varying with x and y as $\displaystyle V = \dfrac{1}{2}(y^2-4x)$ volt. The field at $x = 1 m , y = 1 m$, is :

1. $\displaystyle 2\hat{i}+\hat{j} \ Vm^{-1}$

2. $\displaystyle -2\hat{i}+\hat{j} \ Vm^{-1}$

3. $\displaystyle 2\hat{i}-\hat{j} \ Vm^{-1}$

4. $\displaystyle -2\hat{i}+2\hat{j} \ Vm^{-1}$

Correct Option: C
Explanation:

$\displaystyle E _x = -\dfrac{dV}{dx} = -\dfrac{1}{2} [-4] = 2$

$\displaystyle E _y = -\dfrac{dV}{dy} = -\dfrac{1}{2}[2y] = - y = -1$

$\displaystyle \therefore \vec {E} = E _x \hat{i}+E _y \hat{j}=2\hat{i}-1\hat{j}$

The electric potential $V$ at any point $(x,y,z)$ in space is given by $V=4x^2$ volt. The electric field at $(1,0,2)$m in $Vm^{-1}$ is

1. $8$, along the positive x-axis

2. $8$, along the negative x-axis

3. $16$, along the x-axis

4. $16$, along the z-axis

Correct Option: B

An electric field is expressed as $\displaystyle \vec{E} = 2\hat{i} + 3 \hat{j}$. Find the potential difference $(V _A - V _B)$ between two points $A$ and $B$ whose position vectors are given by $\displaystyle \vec r _A = \hat{i} + 2\hat{j}$ and $\displaystyle \vec r _B = 2\hat{i} + \hat{j}+3\hat{k}$ :

1. $-1 V$

2. $1 V$

3. $2 V$

4. $3 V$

Correct Option: A
Explanation:

$dV=-\vec E.\vec dx-\vec E.\vec dy$
$\Delta V=-\int Edx-\int Edy$
$\displaystyle V _B-V _A = -(\int _{1}^{2}2dx+\int _{2}^{1}3dy)$

$\displaystyle =-[2(2-1)+3(1-2)]$
$\displaystyle =-[2-3] = 1 V$
Hence, $V _A-V _B = -1 V$

An infinite nonconducting sheet of charge has a surface charge density of $10^{-7}\ C/m^2$. The separation between two equipotential surfaces near the sheet whose potential differ by $5\ V$ is

1. $0.88\ cm$

2. $0.88\ mm$

3. $0.88\ m$

4. $5\times 10^{-7}\ m$

Correct Option: B

The electric potential V is given as a function of distance by $V=(5x^2+10x-4)volt$, where x is in metre. Value of electric field at $x=1m$ is :

1. $-23 V/m$

2. $11 V/m$

3. $6 V/m$

4. $-20 V/m$

Correct Option: D
Explanation:

Given, $V=5x^2+10x-4$

or $\dfrac{dV}{dx}=10x+10$
The field, $E=-\dfrac{dV}{dx}=-(10x+10)$

At $x=1, E=-(10+10)=-20 V/m$
So option D is correct.

The potential at a point x (measured in $\mu m$) due to some charges situated on the x-axis is given by $V(x)=20/(x^2-4)volt$
The electric field E at $x=4\mu m$ is given by :

1. $(10/9)volt /\mu m$ and in the $+ve$ x direction

2. $(5/3)volt /\mu m$ and in the $-ve$ x direction

3. $(5/3)volt /\mu m$ and in the $+ve$ x direction

4. $(10/9)volt /\mu m$ and in the $-ve$ x direction

Correct Option: A
Explanation:

Given, $\displaystyle V(x)=\dfrac{20}{x^2-4}$

Electric field , $\displaystyle E=-\dfrac{dV}{dx}=-\dfrac{-20}{(x^2-4)^2}(2x)=\dfrac{40x}{(x^2-4)^2}$

At $\displaystyle x= 4 \mu m, E=\dfrac{40(4)}{(4^2-4)^2}=\dfrac{160}{144}=(10/9) volt/\mu m$

Positive sign indicates that $\vec{E}$ is in the +ve x-direction.

A and B are two points in an electric field. If the work done in carrying $4.0 C$ of electric charge from A to B is $16.0 J$, the potential difference between A and B is :

1. $zero$

2. $2.0 V$

3. $4.0 V$

4. $16.0 V$

Correct Option: C
Explanation:

The work done, $W _{A\rightarrow B}=q\int _{V _A}^{V _B}dV=q(V _B-V _A)$
or $16=4(V _B-V _A) \Rightarrow V _B-V _A=4 V$

Determine the electric field strength vector if the potential of the field depends on x, y coordinates as $V = a (x^2 - y^2)$, where a is a constant.

1. $\vec{E} = - 2a(x\widehat{i} - y\widehat{j})$

2. $\vec{E} = - a(x\widehat{i} - y\widehat{j})$

3. $\vec{E} = - \dfrac{a(x\widehat{i} - y\widehat{j})}{2}$

4. $\vec{E} = - \dfrac{a(x\widehat{i} - y\widehat{j})}{4}$

Correct Option: A
Explanation:

$\vec E = - \triangledown V$   ( negative of gradient of V)

$\vec E = - (\dfrac{dV}{dx} \hat i + \dfrac{dV}{dy} \hat j )=-a(2x \hat i +2y \hat j )$

Determine the electric field strength vector if the potential of the field depends on x, y coordinates as $V = axy$ , where $a$ is a constant.

1. $\vec{E} = -a(y\widehat{i} + y\widehat{j})$

2. $\vec{E} = -a(x\widehat{i} + y\widehat{j})$

3. $\vec{E} = -a(x\widehat{i} + x\widehat{j})$

4. $\vec{E} = -a(y\widehat{i} + x\widehat{j})$

Correct Option: D
Explanation:

$\vec E = - \triangledown V$   ( negative of gradient of V)

$\vec E = - (\dfrac{dV}{dx} \hat i + \dfrac{dV}{dy} \hat j )=-a(y \hat i + x \hat j )$

The electric potential existing in space is $V(x, y, z) = A (xy+ yz + zx)$. Find the expression for the electric field :

1. $-A{(x + Z) \widehat{i} + (y + Z) \widehat{j} + (x + y) \widehat{k}}$

2. $-A{(y + Z) \widehat{i} + (x + Z) \widehat{j} + (x + y) \widehat{k}}$

3. $-Ax{ \widehat{i} +y \widehat{j} + Z\widehat{k}}$

4. $-A{(x+y) \widehat{i} + (x + y) \widehat{j} + (x + y-2Z) \widehat{k}}$

Correct Option: B
Explanation:

$\vec E = - \triangledown V = -A[(y+z) \hat i + ( z+x) \hat j +( y+x) \hat k ] V/m$

At a certain distance from a point charge, the field intensity is 500 V/m and the potential is 3000 V. The distance and the magnitude of the charge respectively are :

1. 6 m and 6 $\mu$C

2. 4 m and 2 $\mu$C

3. 6 m and 4 $\mu$C

4. 6 m and 2 $\mu$C

Correct Option: D
Explanation:

The electric field at distance d due to point charge q is $E=kq/d^2$ and potential $V=kq/d$

so, $E=V/d$ or $d=\dfrac{V}{E}=\dfrac{3000}{500}=6 m$

since, $V=kq/d$

or $3000=9\times 10^9\times \dfrac{q}{6}$

or $q=2\times 10^{-6} C=2 \mu C$

In a certain region of space, the electric potential is $V (x, y, z) = Axy - Bx^2$ $+Cy$, where $A, B\ and\ C$ are positive constants. Calculate the $x, y\ and\ z$ components of the electric field.

1. $E _x = - Ax, E _y = -Ay, E _z = 0$

2. $E _x = - Ax + 2Bx, E _y = -Ay -C, E _z = 0$

3. $E _x = - Ay + 2Bx, E _y = -Ax -C, E _z = 0$

4. $E _x = - Ay, E _y = -Ax, E _z = 0$

Correct Option: C
Explanation:

Here, $V(x,y,z)=Axy-Bx^2+Cy$

So, $E _x=-\dfrac{V}{dx}=-[Ay-2Bx]=2Bx-Ay$;

$E _y=-\dfrac{dV}{dy}=-[Ax+C]=-Ax-C$ and

$E _z=-\dfrac{dV}{dz}=0$

Potential difference between centre and surface of the sphere of radius R and uniform volume charge density $\rho$ within it will be :

1. $\displaystyle \dfrac{\rho R^2}{6 \varepsilon _0}$

2. $\displaystyle \dfrac{\rho R^2}{4 \varepsilon _0}$

3. $\displaystyle \dfrac{\rho R^2}{3 \varepsilon _0}$

4. $\displaystyle \dfrac{\rho R^2}{2 \varepsilon _0}$

Correct Option: A
Explanation:

Using Gauss's law the electric field inside the sphere , $E.4\pi r^2=\dfrac{q}{\epsilon _0}=\rho\dfrac{(4/3)\pi r^3}{\epsilon _0}$

or $E=\dfrac{\rho r}{3\epsilon _0}$

Potential difference between surface and center is $V=-\int _R^0 E.dr=-\int _R^0 \dfrac{\rho r}{3\epsilon _0} dr=\dfrac{\rho R^2}{6\epsilon _0}$

A uniform electric field exists in x-y plane. The potential of points A (-2m, 2m), B(+2m, 2m) and C(2m, 4m) are 4 V, 16V and 12 V respectively. The electric field is :

1. $(4\widehat{i} + 5 \widehat{j}) V/m$

2. $(3\widehat{i} + 4 \widehat{j}) V/m$

3. $-(3\widehat{i} + 4 \widehat{j}) V/m$

4. $(3\widehat{i} - 4 \widehat{j}) V/m$

Correct Option: D
Explanation:
Let equation of potential be $ax+by+c$ where $(x,y)$ are the co-ordinates of the point in $x, y$ plane.
so, $a(2)+b(2)+c=4$
$a(-2)+b(2)+c=16$
$a(2)+b(4)+c=12$
Solving above equations, we get $a=-3;b=4;c=2$
so equation of potential is $V=-3x+4y+2$
Now the electric field is $\vec E=(-dV/dx)\vec i+(-dV/dy)\vec j=3\vec i-4\vec j$

In a certain region of space, the electric potential is $V (x, y, z) = Axy - Bx^2$ $+Cy$, where $A, B\ and\ C$ are positive constants. At which points is the electric field equal to zero?

1. $x = +C/A, y = +BC/A$ $^2$, any value of $z$

2. $x = +C/A, y = +2BC/A$ $^2$, any value of $z$

3. $x = -C/A, y = -2BC/A$ $^2$, $z=0$

4. $x = -C/A, y = -2BC/A$ $^2$, any value of $z$

Correct Option: D
Explanation:

$\vec E = - \triangledown V = - [ ( Ay - 2Bx) \hat i + ( Ax + C) \hat j ]$

$\vec E = 0$  at,

$E _y = 0 \Rightarrow Ax + C= 0 \Rightarrow x = -C/A$

$E _x= 0 \Rightarrow Ay - 2Bx= Ay + 2BC/A =0 \Rightarrow y = -2BC/A^2$

$E _z = 0$  everywhere .

The electric potential existing in space is $V(x, y, z) = A (xy+ yz + zx)$. If A is $10$ SI units, find the magnitude of the electric field at $(1 m, 1 m, 1 m)$ :

1. $20 \sqrt 2$ N/C

2. $20 \sqrt 3$ N/C

3. $10 \sqrt 3$ N/C

4. $20$ N/C

Correct Option: B
Explanation:

$\vec E = - \triangledown V = -A[(y+z) \hat i + ( z+x) \hat j +( y+x) \hat k ] = -10[2 \hat i + 2 \hat j + 2 \hat k ] = 20\sqrt{3} N/C$

The electric potential at a point (x, y) in the x-y plane is given by V = - Kxy. The field intensity at a distance r in this plane, from the origin is proportional to :

1. $r^2$

2. $r$

3. $1/r$

4. $1/r^2$

Correct Option: B
Explanation:
Let the $x-y$ coordinates of the point at distance $r$ from the origin be given as $x=rcos\theta$, $y=rsin\theta$
Potential is given as $V=-Kxy$
Now, Electric field intensity is $\vec E=(-dV/dx)\vec i+(-dV/dy)\vec j=K(y\vec i+x\vec j)=K(rsin\theta\vec i+rcos\theta\vec j)=Kr(sin\theta\vec i+cos\theta\vec j)\propto r$
So electric field potential is proportional to $r$.

Find out the relationship between the electric field and electric potential include which of the following statement?

I. If the electric field at a certain point is zero, then the electric potential at the same point is also zero.

II. The electric potential is inversely proportional to the strength of the electric field.

III. If the electric potential at a certain point is zero, then the electric field at the same point is also zero.

1. I only

2. II only

3. I and II only

4. I and III only

5. None of the above

Correct Option: E
Explanation:

If electric field at some point is zero, it is not necessary for the electric potential to be the same. Consider the mid point of line joining to equal charges of same sign. Field there is zero, but potential is finite and positive.

Electric field strength is given by $E=\dfrac{kQq}{r^2}$
Electric potential is given by $V=\dfrac{dQq}{r}$
Clearly they are not inversely proportional to each other.

If electric potential at some point is zero, it is not necessary for the electric field to be the same. Consider the mid point of line joining to equal charges of opposite signs. Potential there is zero, but electric field exists.

When negative charges are kept in electric field then negative charges are accelerated by electric fields toward points:

1. at lower electric potential

2. at higher electric potential

3. where the electric field is zero

4. where the electric field is weaker

5. where the electric field is stronger

Correct Option: B
Explanation:

We know that , $F=-\Delta U$, change in potential energy.

As charge is negative so, $F=-qE=-\Delta U$ or $\Delta U=qE$
So we will get change in potential as positive and hence the electric field should be towards  points at higher potential.

An electric field (in $V/m$) is given by $E=10x^3$. Determine the potential difference, in volts, between $x=0m$ and $x=3m$.

1. $202.5$

2. $100$

3. $20$

4. $250$

Correct Option: A
Explanation:

We know that, $E=-\dfrac{dV}{dx}$

$V=-\int _0^3 Edx=-\int _0^3 10x^3 dx=-10\times \dfrac{3^4}{4}=202.5$

In the direction of electric field, the electric potential:

1. Decreases

2. Increases

3. Remains unchanged

4. Becomes zero

Correct Option: A
Explanation:

$dV=-E.dr$, so the electric potential $V$ decreases continuously as we move along the direction of the electric field

The most appropriate relationship between electric field and electric potential can be described as
($C$ is an arbitrary path connecting the point with zero potential infinity)

1. $V _E = -\int _C E.dl$

2. $E _V = -\int _C V.dl$

3. $V _E = -\int E.dl$

4. $E _V = -\int E.dl$

Correct Option: A
Explanation:
Let $\Delta V=V _{B}-V _{A}$ be the electrostatic potential energy difference between any two points A and B in the electric field. The electric potential difference between points A and B is given by:
$\Delta V= V _{B}-V _{A}=\dfrac{\Delta V}{q _{0}}=\dfrac{V _{B}-V _{A}}{q _{0}}$
$\Delta V=V _{B}-V _{A}=-q _{0}\int _{A}^{B}\overrightarrow{E}.\overrightarrow{dl}$
$\Delta V=\dfrac{\Delta V}{q _{0}}=-\int _{A}^{B}\overrightarrow{E}.\overrightarrow{dl}$      ...(i)
If C is an arbitrary path connecting the point with zero potential at infinity. Then, equation (i) becomes:
$V=-\int _{C}^{ }E.dl$

The potential in a certain region of space is given by the function $xy^2z^3$ with respect to some reference point. Find the y-component of the electric field at $(1, -3, 2)$.

1. $48 \hat j$

2. $48 \hat i$

3. $-48 \hat i$

4. $-48 \hat j$

Correct Option: A
Explanation:

given potential $=x{ y }^{ 2 }{ z }^{ 2 }$

we have to find the y-component of electric field at $(1,-3,2)$
here, we use the relation b/w electric field and the potential $E=\cfrac { -dv }{ dr }$
To find y- c ordinate of electric field, we differential function of V and y.
so,${ E } _{ y }=\cfrac { -dv }{ dy } =\cfrac { -d }{ dy } (x{ y }^{ 2 }{ z }^{ 3 })$
${ E } _{ y }=-2x{ y }{ z }^{ 3 }$
To find electric field at pt. $(1,-3,2)$
we substitute for $x=1$
$y=-3$
$z=2$
we get${ E } _{ y }=-2\left( 1 \right) \left( -3 \right) { \left( 2 \right) }^{ 3 }$
$=48$
Hence the answer is $48\hat {j}$
so, the correct answer is option $(a).$

If $4\times 10^{20}eV$ of energy is required to move a charge of $0.25$ coulomb between two points, the p.d between them is:

1. $256\ V$

2. $512\ V$

3. $123\ V$

4. $215\ V$

Correct Option: A
Explanation:

$4\times 10^{20}eV=4\times 10^{20}\times 1.6\times 10^{-19}=64 J$

So $E=64=V\times Q=0.25V$
$V=256 V$

A uniform electric field  of $12$ $V/m$ is along the positive $x$ direction. Determine the potential difference in volts, between $x=0m$ and $x=3m$.

1. $-27$ $V$

2. $-36$ $V$

3. $27$ $V$

4. $36$ $V$

Correct Option: B
Explanation:
As we know, relation between uniform electric field strength $\&$ potential difference between two point a distance d
$V=-Ed \; \Rightarrow \; V=-(12)(3)=-36 \; Volts$

In the direction of electric field, the electric potential:

1. decreases

2. increases

3. remains uncharged

4. becomes zero

Correct Option: A
Explanation:

In the direction of electric field the electric potential decreases. This is because electric potential is the work done against the direction of electric field.

Variation of potential V with distance r in electric field of E$=0$ is?

1. $V\propto \displaystyle\frac{1}{r}$

2. $V\propto r$

3. $V\propto \displaystyle\frac{1}{r^2}$

4. $V=$ constant

Correct Option: D
Explanation:

Potential difference between two points is given by    $\Delta V = -E.r$
Given :  $E = 0$
$\implies \ \Delta V = 0$
$\implies \ V =$ constant

The electric potential decreases uniformly from V to -V along X-axis in a coordinate system as we moves from a (-$x _0$, 0) to ($x _0$, 0), then the electric field at the origin.

1. must be equal to $\dfrac{V}{x _0}$;

2. may be equal to $\dfrac{V}{x _0}$;

3. must be greater than $\dfrac{V}{x _0}$;

4. may be less than $\dfrac{V}{x _0}$;

Correct Option: C

A region, the potential is given by V=-{5x + 5y + 5z}, where V is in volts and x, y, z are in meters. The intensity of the electric field is:

1. $2$ V/m

2. $3\sqrt3$ V/m

3. $2\sqrt2$V/m

4. $5\sqrt3$ V/m

Correct Option: B

A copper ball of radius 1 cm work function 4.47 eV is irradiated with ultraviolet radiation of wavelength $2500\mathring { A }$. The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be finite value of the potential on the ball. The charge acquired by the ball is :

1. $5.5\times { 10 }^{ -13 }C$

2. $7.5\times { 10 }^{ -13 }C$

3. $4.5\times { 10 }^{ -12 }C$

4. $2.5\times { 10 }^{ -11 }C$

Correct Option: A
Explanation:

From photo electric effect equation :

]
$h\nu=h\nu _{0} + K.E _{max}$

so Maximum kinetic energy will be

$K.E _{max}= \dfrac{hc}{\lambda} - h\nu _{0}$

putting the given values in the above equation

$K.E _{max} = e\times V$

so V will be

$V= \dfrac{k\times Q}{r}$

:: $q = 5.5\times 10^{-13} C$

Two infinite, parallel, non-conducting sheets carry equal positive charge density $\sigma$. One is placed in the yz plane at $x=0$ and the other at distance $x=a$. Take potential $V=0$ at $x=0$. Then,

1. for $0\leq x \leq a$, potential $V _x=0$

2. for $x\geq a$, potential $V _x=-\frac {\sigma}{\epsilon _0}(x-a)$

3. for $x\geq a$, potential $V _x=\frac {\sigma}{\epsilon _0}(x-a)$

4. for $x\leq 0$ potential $V _x=\frac {\sigma}{\epsilon _0}x$

Correct Option: A,B,D
Explanation:

Now , Since both are infinite plates and carry same charge density therefore, electric field between them will be equal to zero.
Now, potential will be constant between them and at $x=0; V=0$ and V = constant between the plates.
Therefore, V=0 between the plates means $0\le x\le a$
Now electric field beyond $x=a$ is $2\times \sigma/2\epsilon _o=\sigma/\epsilon _o$
We know that,
$V=-\int _{ a }^{ x }{ \overrightarrow { E } .\overrightarrow { dx } }$
$V=-E(x-a)$
$V=-\sigma(x-a)/\epsilon _o$

and for $x<0$
$V=-\int _{ x }^{ 0 }{ \overrightarrow { E } .\overrightarrow { dx } }$
$V=Ex$
$V=\sigma x/\epsilon _o$
option (A)(B)(D) are correct.

Electric potential $'v'$ in space as a function of co-ordinates is given by, $v=\cfrac{1}{x}+\cfrac{1}{y}+\cfrac{1}{z}$. Then the electric field intensity at $(1,1,1)$ is given by :

1. $-(\hat { i } +\hat { j } +\check { k } )$

2. $\hat { i } +\hat { j } +\check { k }$

3. zero

4. $\cfrac{1}{\sqrt 3}(\hat { i } +\hat { j } +\check { k } )$

Correct Option: B
Explanation:
The electric field , $\vec{E}=-\vec{\nabla}V=-\left[\dfrac{\partial V}{\partial x}\hat i+\dfrac{\partial V}{\partial y}\hat j+\dfrac{\partial V}{\partial z}\hat k\right]=\dfrac{1}{x^2}\hat i+\dfrac{1}{y^2}\hat j+\dfrac{1}{z^2}\hat k$
at $(1,1,1) \Rightarrow \vec{E}=\hat i+\hat j+\hat k$

The electrostatic potential inside a charged spherical ball is given by $\phi=ar^2+b$, where r is the distance from the centre and a, b are constant. Then the charge density inside the ball is :

1. $-6 a \epsilon _0r$

2. $-24\pi a \epsilon _0r$

3. $-6 a \epsilon _0$

4. $-24 \pi a \epsilon _0$

Correct Option: C
Explanation:
Electric filed , $E=-\dfrac{d\phi}{dr}=-2ar$
By Gauss's law, $E.4\pi r^2=\dfrac{q _{in}}{\epsilon _0}$
$\Rightarrow q _{in}=(-2ar)4\pi r^2 \epsilon _0=-8\pi \epsilon _0 ar^3$
Now $\dfrac{dq _{in}}{dr}=-24\pi \epsilon _0 ar^2$ and $V=\dfrac{4}{3}\pi r^3, \dfrac{dV}{dr}=4\pi r^2$
Charge density , $\rho=\dfrac{dq _{in}}{dV}=\dfrac{dq _{in}}{dr}\times \dfrac{dr}{dV}=(-24\pi \epsilon _0 ar^2)\times \dfrac{1}{4\pi r^2}=-6 \epsilon _0 a$

An electric field is given by $\vec E = (y \hat i + \hat x) NC^{-1}$. Find the work done (in $J$) by the electric field in moving a $1\ C$ charge from $\vec r _A = (2 \hat i + 2 j) m$ to $\vec r _B = (4 \hat i + \hat j) m$

1. $0\ J$

2. $-2\ J$

3. $2\ J$

4. $4\ J$

Correct Option: A
Explanation:

Work done , $W=\int \vec{F}.\vec{dr}$

Here electrostatic force , $\vec{F}=q\vec{E}=q(y\hat i+x\hat j)$
$\vec{F}.\vec{dr}=q(y\hat i+x\hat j).(dx\hat i+dy\hat j)=q(ydx+xdy)=d(xy)$  as $q=1 C$

Now $W=\int _{2,2}^{4,1}d(xy)=[xy] _{2,2}^{4,1}=4\times 1-2\times 2=4-4=0$

If the electrostatic potential is given by $\phi =\phi _0(x^2+ y^2 + z^2)$ where $\phi _0$ is constant, then the charge density of the given potential would be :

1. $0$

2. $-6\phi _0\varepsilon _0$

3. $-2\phi _0\varepsilon _0$

4. $\dfrac{-6\phi _0}{\varepsilon _0}$

Correct Option: B
Explanation:

$\overrightarrow{E} = -\triangledown \phi$
$\overrightarrow{\triangledown}.\overrightarrow{E} = \rho/\epsilon _0$
Now, $\phi = \phi _0 (x^2 + y^2 + z^2) \Rightarrow \overrightarrow{E} = -2\phi _0 ( \hat{i}+\hat{j}+\hat{k} ) \Rightarrow \rho = -6\phi _0 \epsilon _0$

Electric field in a region is given as $\bar{E}=x\hat{i}+2y\hat{j}+3\hat{k}$. In this region point A(3,3,1) and point B (4,2,1) are there. The magnitude of work done by the electric field, if 2 coulomb charge is moved from A to B. All values are in SI units:

1. 3

2. 4

3. 5

4. 6

Correct Option: A
Explanation:

Given, $\vec{E}=x\hat{i}+2y\hat{j}+3\hat{k}$ and $q=2 C$
Work done, $W=\int^B _Aq\vec{E}.\vec{dr}=q\int^B _A(x\hat{i}+2y\hat{j}+3\hat{k}).(dx\hat{i}+dy\hat{j}+dz\hat{k})$
or,$W=2\int^{(4,2,1)} _{(3,3,1)}xdx+2ydy+3dz=2[\frac{16-9}{2}+(4-9)+3(1-1)]=7-10=-3$
Magnitude of work done$=|W|=3$

Find the magnitude of the force on a charge of $12\mu C$ placed at point where the potential gradient has a magnitude of $6\times 10^{5}V\ m^{-1}$

1. $5.20\ N$

2. $7.20\ N$

3. $6.20\ N$

4. $8.20\ N$

Correct Option: B
Explanation:

Potential gradient is nothing but the rate of change of electric potential with position and it is equal to electric field at that point.

$\dfrac{dV}{dl}=E$=electric field

$\implies E=6\times 10^5Vm^{-1}$

Force on charge $=F=qE=12\times 10^{-6}\times 6\times 10^5$

$\implies F=7.2N$

The most appropriate relationship between electric field and electric potential is given by

1. $E = - \nabla V _E$

2. $V _E = - \nabla E$

3. $E = \nabla V _E$

4. $V = - \nabla E$

Correct Option: A
Explanation:

$\vec{E}=-\dfrac{\partial V}{dx}\hat{i}-\dfrac{\partial V}{dy}\hat {j}-\dfrac{\partial V}{dz}\hat{k}$

And, we know that $\nabla=\dfrac{\partial}{dx}+\dfrac{\partial}{dy}+\dfrac{\partial}{dz}$

Hence, we get $\vec{E}=-\nabla V$

Electrostatic potential energy of a shell of radius $10cm.$ When $10C$ charge is distributed over its surface.

1. $4.5 \times {10^{12}}J$

2. $5.4 \times {10^8}J$

3. $4.5 \times {10^9}J$

4. $5.4 \times {10^6}J$

Correct Option: A
Explanation:

$U = \dfrac{{k\,Q \cdot Q}}{{2R}}$

$= \dfrac{{9 \times {{10}^9} \times 10 \times 10}}{{2 \times 0.1}}$
$U = 4.5 \times {10^{12}}J$

Two charges $+Q$ and $-2Q$ are located at points $A$ and $B$ on a horizontal line as shown in the diagram.
The electrical field is zero at a point which is located at finite distance :

1. On the perpendicular bisector of $AB$

2. Left of $A$ on the line

3. Between $A$ and $B$ on the line

4. Right of $B$ on the line

Correct Option: B
Explanation:

$E _1$ = Electrical field due to $+Q$
$E _2$ = Electrical feild due to $-2Q$
There resultant is $0$ at this point

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