Tag: relation between electric field and electric potential

Questions Related to relation between electric field and electric potential

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