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

Two infinite, parallel, non-conducting sheets carry equal positive charge density $\sigma$. One is placed in the yz plane at $x=0$ and the other at distance $x=a$. Take potential $V=0$ at $x=0$. Then,

  1. for $0\leq x \leq a$, potential $V _x=0$
  2. for $x\geq a$, potential $V _x=-\frac {\sigma}{\epsilon _0}(x-a)$
  3. for $x\geq a$, potential $V _x=\frac {\sigma}{\epsilon _0}(x-a)$
  4. for $x\leq 0$ potential $V _x=\frac {\sigma}{\epsilon _0}x$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

Now , Since both are infinite plates and carry same charge density therefore, electric field between them will be equal to zero.
Now, potential will be constant between them and at $x=0; V=0$ and V = constant between the plates.
Therefore, V=0 between the plates means $0\le x\le a$
Now electric field beyond $x=a$ is $2\times \sigma/2\epsilon _o=\sigma/\epsilon _o$
We know that,
$V=-\int _{ a }^{ x }{ \overrightarrow { E } .\overrightarrow { dx }  } $
$V=-E(x-a)$
$V=-\sigma(x-a)/\epsilon _o$

and for $x<0$
$V=-\int _{ x }^{ 0 }{ \overrightarrow { E } .\overrightarrow { dx }  } $
$V=Ex$
$V=\sigma x/\epsilon _o$
option (A)(B)(D) are correct.

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

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 

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A copper ball of density $8.6 g cm^{-3}$, 1 cm is diameter is immersed in oil of density $0.8cm^{-3}$ . if the ball remains suspended in oil in a uniform electric field of intensity $36000 NC^{-1} $acting in upward direction, what is the charge on the ball ? 

  1. $1.1 \mu C$
  2. $4.2 \mu C$
  3. $2.4 \mu C$
  4. $3.7 \mu C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the ball to be suspended: electric force = weight - buoyant force. qE = ρ_ball V g - ρ_oil V g. q = (ρ_ball - ρ_oil)Vg/E. Volume of sphere = (4/3)πr³ = (4/3)π(0.5 cm)³. After unit conversions and calculation, q ≈ 1.1 μC. Option A is correct.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A solid ball of radius R has a charge density p given by $p=p _0(1 -r/R)$ for $ 0 \leq r \leq R.$ The electric field outside the ball is:

  1. $\dfrac{p _0R^3}{\epsilon _0r^2}$
  2. $\dfrac{p _0R^3}{12\epsilon _0r^2}$
  3. $\dfrac{4p _0R^3}{3\epsilon _0r^2}$
  4. $\dfrac{3p _0R^3}{4\epsilon _0r^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$According\, to\, question..................... \ q=\int  _{ 0 }^{ R }{ pdv }=\int  _{ 0 }^{ R }{ po }\left( { 1-\dfrac { r }{ R }  } \right) \, .\, 4\pi { r^{ 2 } }drr \ \Rightarrow q=po.4\pi \, \left[ { \int  _{ 0 }^{ R }{ { r^{ 2 } }dr-\int  _{ 0 }^{ R }{ \dfrac { { { r^{ 3 } } } }{ R } dr } } } \right]  \ \Rightarrow q=po.4\pi \left( { \dfrac { { { R^{ 3 } } } }{ 3 } -\dfrac { { { R^{ 3 } } } }{ 4 }  } \right) =po\left( { 4\pi  } \right) \dfrac { { { R^{ 3 } } } }{ { 12 } }  \ \, \, \, \, \, \, \therefore \, \, \, q=\dfrac { { po.\pi { R^{ 3 } } } }{ 3 }  \ E=\dfrac { 1 }{ { 4\pi { \in _{ 0 } } } } \times \, \dfrac { q }{ { { r^{ 2 } } } } =\dfrac { 1 }{ { 4\pi { \in _{ 0 } } } } \, \times \dfrac { { po\, .\pi { R^{ 3 } } } }{ { 3{ r^{ 2 } } } } =\dfrac { { po\, .{ R^{ 3 } } } }{ { 12{ \in _{ 0 } }{ r^{ 2 } } } }  \ \, \, \, \therefore \, \, \, \, E=\dfrac { { po\, .{ R^{ 3 } } } }{ { 12{ \in _{ 0 } }{ r^{ 2 } } } }  \ there\, for\, the\, correct\, option\, is\, B.$

Multiple choice polarisation of light polarisation wave optics optics physics

Plane of polarisation is:

  1. the plane in which vibrations of the electric vector takes place

  2. a plane perpendicular to the plane in which vibrations of the electric vector takes place

  3. perpendicular to the plane of vibration

  4. horizontal plane
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$ \bf{Plane\ of\ Vibrations} $ is the plane in which the vibrations of electric vector of plane polarized light are present. 


$ \bf{Plane\ of\ Polarization} $ is the plane perpendicular to the plane of vibration and in this plane, the vibrations of the electric vector are absent.

Hence, the correct answers are OPTIONS B and C. 

Multiple choice polarisation of light polarisation wave optics optics physics

A polaroid making an angle ${ 60 }^{ \circ  }$ with electric vector then intensity reduced by a factor of:-

  1. $\dfrac { 1 }{ 4 } $
  2. $\dfrac { 3}{ 4 } $
  3. $\dfrac { 1 }{ 2 } $
  4. $\dfrac { 1 }{ 3} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a polarizer makes an angle of 60 degrees with the electric vector of polarized light, the intensity is reduced by a factor of cos^2(60) = (1/2)^2 = 1/4.

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding . The charges on the two rings are + q and -q . The potential difference between the centres of the two rings is

  1. $\frac {{ Q.R }\quad} {\quad { 4\pi }{ \varepsilon } _{ 0 }{ d }^{ 2 }}$
  2. $\frac { Q }{ 2\pi { \varepsilon } _{ 0 } } [\frac { 1 }{ R } -\frac { 1 }{ \sqrt { { R }^{ 2 }+{ d }^{ 2 } } } ]$
  3. $\frac { Q }{ { 4\pi \varepsilon } _{ 0 } } [\frac { 1 }{ R } -\frac { 1 }{ \sqrt { { { R }^{ 2 } }+{ { d }^{ 2 } } } } ]$

  4. 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Two thin wire ring each having a radius $=R$

distance $=d$
two rings are $+q$ and $-q$.
$PD=\dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \dfrac { QR }{ { d }^{ 2 } } $
Now in this question, we get the solution.