Tag: relation between electric field and electric potential

Questions Related to relation between electric field and electric potential

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 :

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

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

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

  3. zero

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

Show answer
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 :

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

  1. $-6 a \epsilon _0r$

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

  3. $-6 a \epsilon _0$

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

Show answer
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 $\vec r _B = (4 \hat i + \hat j) m$

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

  1. $0\ J$

  2. $-2\ J$

  3. $2\ J$

  4. $4\ J$

Show answer
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 :

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

  1. $0$

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

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

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

Show answer
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:

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

  1. 3

  2. 4

  3. 5

  4. 6

Show answer
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 $6\times 10^{5}V\ m^{-1}$

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

  1. $5.20\ N$

  2. $7.20\ N$

  3. $6.20\ N$

  4. $8.20\ N$

Show answer
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 $=F=qE=12\times 10^{-6}\times 6\times 10^5$

$\implies F=7.2N$

Answer-(B)

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

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

  1. $E = - \nabla V _E$

  2. $V _E = - \nabla E$

  3. $E = \nabla V _E$

  4. $V = - \nabla E$

Show answer
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 $\vec{E}=-\nabla V$

Answer-(A)

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

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

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

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

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

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

Show answer
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 :

physics electrostatics field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  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

Show answer
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