Tag: relation between electric field and electric potential

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Questions Related to relation between electric field and electric potential

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

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

  3. zero
  4. $\cfrac{1}{\sqrt 3}(\hat { i } +\hat { j } +\check { k } )$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $(1,1,1) \Rightarrow \vec{E}=\hat i+\hat j+\hat k$
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 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. $-24 \pi a \epsilon _0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 , $\rho=\dfrac{dq _{in}}{dV}=\dfrac{dq _{in}}{dr}\times \dfrac{dr}{dV}=(-24\pi \epsilon _0 ar^2)\times \dfrac{1}{4\pi r^2}=-6 \epsilon _0 a$
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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. $4\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $W=\int _{2,2}^{4,1}d(xy)=[xy] _{2,2}^{4,1}=4\times 1-2\times 2=4-4=0$

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 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. $\dfrac{-6\phi _0}{\varepsilon _0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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$=|W|=3$

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

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. $8.20\ N$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)

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 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. $V = - \nabla E$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

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

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. $5.4 \times {10^6}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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

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