Tag: relation between electric field and electric potential

Questions Related to relation between electric field and electric potential

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 ${ r }^{ 2 }$