Physics

Electrostatics

299 Questions

Electrostatics deals with electric charges, fields, and potentials at rest. It is a crucial topic for physics sections in engineering and civil services competitive examinations. Review these questions to build a strong understanding of Coulomb law, electric dipoles, Gauss law, and electric flux.

Electric field and potentialElectric dipole momentGauss Law applicationsCoulomb force calculationsCharge distribution on spheresEquipotential surfaces

Electrostatics Questions

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
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 $\overrightarrow{E} = (-8)\hat{i} \,V/m$.
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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. $\dfrac{10}{3}\ volt$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

E = -dV/dx. The potential changes by 40 V over 4 m. The average field is 10 V/m. Since it decreases uniformly, the field is constant at 10 V/m everywhere in that interval.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

$64$ 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. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume is conserved: 64 * (4/3)pi*r^3 = (4/3)pi*R^3, so R = 4r. Potential V = kQ/r. Q_new = 64q. V_new = k(64q)/(4r) = 16 * (kq/r) = 16V.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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. $\dfrac{1}{20}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Electric field $=10N/C$
Vertically downward direction electric potential as one goes through $h=50cm$ $=50\times { 10 }^{ -2 }m$
Now, $V=E/d$
$=10/50\times { 10 }^{ -2 }=\dfrac { 100 }{ 5 } =20J$
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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. $-(3\hat i + 2\hat j + 6\hat k )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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


Formula,

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

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

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

$=3i+2j+6k$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 ${ r }^{ 2 }$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric field E between two parallel plates is given by E = V/d. Here, V = 10 V and d = 1 cm = 0.01 m, so E = 10 / 0.01 = 1000 N/C.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
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., $m _1 \times m _2=-1$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
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.