Physics

Electrostatics

303 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 applications of gauss's law coulomb's law physics

The magnitude of the electric field on the surface of a sphere of radius $r$ having a uniform surface charge density $\sigma$ is

  1. $\sigma / \epsilon _{0}$
  2. $\sigma / 2\epsilon _{0}$
  3. $\sigma / \epsilon _{0}r$
  4. $\sigma / 2\epsilon _{0}r$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The magnitude of the electric field on the surface of radius $=r$
Charge density $=6$
Then, $E=\dfrac { 6 }{ { \epsilon  } _{ 0 } } $
The electric field is independent of the surface radius.
Multiple choice applications of gauss's law coulomb's law physics

Consider a thin spherical shell of radius $R$ consisting of uniform surface charge density $\sigma$. The electric field at a point of distance $x$ from its centre and outside the shell is

  1. inversely proportional to $\sigma$
  2. directly proportional to ${x}^{2}$
  3. directly proportional to $R$
  4. inversely proportional to ${x}^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
For a thin uniformly charged spherical shell, the field points outside the shell at a distance $x$ from the centre is
$E=\cfrac { 1 }{ 4\pi { \varepsilon  } _{ 0 } } \cfrac { Q }{ { x }^{ 2 } } $
If the radius of the sphere is $R,Q=\sigma 4\pi { R }^{ 2 }$
$\therefore E=\cfrac { 1 }{ 4\pi { \varepsilon  } _{ 0 } } \cfrac { \sigma 4\pi { R }^{ 2 } }{ { x }^{ 2 } } =\cfrac { \sigma { R }^{ 2 } }{ { { \varepsilon  } _{ 0 }x }^{ 2 } } $
This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre
Multiple choice applications of gauss's law coulomb's law physics

Two charged spheres having radii a and b are joined with a wire then the ratio of electric field $\dfrac{E _a}{E _b}$ on their surface is?

  1. a/b

  2. b/a

  3. ba

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When the two spheres are connected by a wire, then both of them acquire the same potential say $V$.


We also know that the electric field on the surface of a sphere $E=\dfrac{Q}{4\pi\epsilon _o r^2}$
and potential on the surface is given by $V=\dfrac{Q}{4\pi\epsilon _or}$

$\implies E=\dfrac{V}{r}$

Here, V is constant , hence  $E\propto \dfrac{1}{r}$

$\implies \dfrac{E _a}{E _b}=\dfrac{b}{a}$

Multiple choice applications of gauss's law coulomb's law physics

Charges $Q _1$ and $Q _2$ are placed inside and outside respectively of an uncharged conducting shell. Their seperation is r.

  1. The force on $Q _1$ is zero.
  2. The force on $Q _1$ is $\displaystyle k \frac{Q _1 Q _2}{r^2}$
  3. The force on $Q _2$ is $\displaystyle k \frac{Q _1 Q _2}{r^2}$
  4. The force on $Q _2$ is zero.
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

As the electric field inside the conducting shell is zero , so the force on the inner charge, $Q _1$ will be zero.
The electric field at outside charge $Q _2$ due to $Q _1$ is $E=k\frac{Q _1}{r^2}$
Force on $Q _2$ is $F=Q _2E=k\frac{Q _1Q _2}{r^2}$

Multiple choice ionic bond atomic structure and chemical bonding chemistry

According to Coulomb's law, the force of attraction (F) between two oppositely charged ions separated by a distance d in air is given by

  1. $F = \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{d^2}$
  2. $F = \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{(r^+ + r^-)^2}$
  3. $F = {4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{d}$
  4. $F = {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{(r^+ + r^-)}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

According to Coulomb's law, the force of attraction (F) between two oppositely charged ions separated by a distance d in air is given by
$F = \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{d^2}$ or $F = \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{(r^+  + r^-)^2}$

Multiple choice physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

An electric dipole of length $20cm$ having $\pm 3\times { 10 }^{ -3 }C$ charge placed at ${60}^{o}$ with respect to a uniform electric field experiences a torque of magnitude $6Nm$. The potential energy of the dipole is

  1. $-2\sqrt{3}J$
  2. $5\sqrt{3}J$
  3. $-2\sqrt {2}J$
  4. $3\sqrt {5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here length of dipole $2a=20cm=20\times { 10 }^{ -2 }m$, Charge $q=\pm 3\times { 10 }^{ -3 }C,\theta ={ 60 }^{ o }\quad $ and torque $\tau =6Nm$
As $\tau =pE\sin { \theta  } $
or $E=\cfrac { \tau  }{ p\sin { \theta  }  } =\cfrac { \tau  }{ q(2a)\sin { \theta  }  } \left( \because p=q(2a) \right) $
$\therefore E=\cfrac { 6 }{ 3\times { 10 }^{ -3 }\times 20\times { 10 }^{ -2 }\times \sin { { 60 }^{ o } }  } =\cfrac { { 10 }^{ 5 } }{ 5\sqrt { 3 }  } N{ C }^{ -1 }$
Potential energy of dipole $U=-pE\cos{\theta}=-q(2a)E\cos{\theta}$
$=-3\times { 10 }^{ -3 }\left( 20\times { 10 }^{ -2 } \right) \cfrac { { 10 }^{ 5 } }{ 5\sqrt { 3 } } \cos { { 60 }^{ o } } =\cfrac { -3\times { 10 }^{ -5 }\times 20\times { 10 }^{ 5 } }{ 5\sqrt { 3 } \times 2 } =-2\sqrt { 3 } J\quad \quad $

Multiple choice physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

An electric dipole has the magnitude of its charge as $q$ and its dipole moment is $p$. It is placed in uniform electric field $E$. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively

  1. $q.E$ and max
  2. $2q.E$ and min.
  3. $q.E$ and min

  4. zero and min.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When the dipole is in the direction of field then net force is $qE+(-qE)=0$
and its potential energy is minimum $=-p.E$
$=-qaE$

Multiple choice physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

An electric dipole of diploe moment $\overrightarrow { p } $ placed in uniform electric field $\overrightarrow { E } $ has minimum potential energy when angle between $\overrightarrow { p } $ and $\overrightarrow { E } $

  1. $\cfrac{\pi}{2}$

  2. zero
  3. $\pi$
  4. $\cfrac{3\pi}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Potential Energy=$ -PE \cos {\theta}$

when 
$ \theta=0 $
Potential Energy=$ -PE $
When
$ \theta=180 $
Potential Energy=$ +PE $
So, Maximum Potential Energy=$ +PE $ at angle $\theta=\pi$

Multiple choice physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

 Two small electric dipoles each of dipole moment pi are situated at $(0, 0, 0)$ and $(r, 0, 0)$. the electric potential at a point $\left( \frac { r } { 2 } , \frac { \sqrt { 3 } r } { 2 } , 0 \right)$ is:

  1. $\frac { p } { 4 \pi \in _ { 0 } r ^ { 2 } }$
  2. $0$
  3. $\frac { p } { 2 \pi \epsilon _ { 0 } r ^ { 2 } }$
  4. $\frac { p } { 8 \pi \epsilon _ { 0 } r ^ { 2 } }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The point (r/2, sqrt(3)r/2, 0) forms an equilateral triangle with the two dipoles at (0,0,0) and (r,0,0). The potential from the first dipole is V1 = (p cos theta1) / (4 pi epsilon0 r1^2) and from the second is V2 = (p cos theta2) / (4 pi epsilon0 r2^2). Summing these at the given coordinates yields the result.

Multiple choice physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

Potential at any point in the electric field produced by a dipole is

  1. $\infty , r$
  2. $\alpha r ^ { 2 }$
  3. $\frac { 1 } { r }$
  4. $\frac { 1 } { r ^ { 2 } }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric potential at a point due to an electric dipole at a distance r is inversely proportional to the square of the distance (V proportional to 1/r^2), unlike a point charge where potential is proportional to 1/r.

Multiple choice physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

A dipole of dipole moment $\overline {\text{p}} $ i s aligned at right angle to electrictric field $\overline {\text{E}} $ . To set it at an angle $\theta $ with E the amount of work done is


  1. $ - {\text{pEcos}}\theta $
  2. $ {\text{pEsin}}\theta $
  3. $ - {\text{pE}}\left( {{\text{sin}}\theta - 1} \right)$
  4. $ - {\text{pE}}\left( {{\text{sin}}\theta + 1} \right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work done in rotating a dipole in an electric field is W = U_final - U_initial. U = -pE cos(theta). Initial angle is 90 degrees (cos 90 = 0). Final angle is theta. W = -pE cos(theta) - 0 = -pE cos(theta).

Multiple choice physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

A electric dipole moment $\vec { p } =\left( 2.0\hat { i } +3.0\hat { j }  \right) \mu C.m$ is placed in a uniform electric field $\vec { E } =\left( 3.0\hat { i } +2.0\hat { k }  \right) \times { 10 }^{ 5 }N{ C }^{ -1 }$

  1. The torque that $\vec { E }$ exerts on $\vec { p }$ is $\left( 0.6\hat { i } -0.4\hat { j } -0.9\hat { k } \right) Nm$
  2. The potential energy of the dipole is $-0.6J$
  3. The potential energy of the dipole is $0.6J$
  4. If the dipole is free to rotate in the electric field, the maximum magnitude of potential energy of the dipole during the rotation is $1.3J$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

$\vec P = (2 \widehat i + 3 \widehat j) \mu cm$.
$\vec E = (0.3 \widehat i + 0.2 \widehat k) N \mu C^{-1}$
$\vec C = \vec P \times \vec E$
$=\begin{vmatrix}\widehat i & \widehat j & \widehat k\ 2 & 3 & 0\0.3  & 0 & 0.2\end{vmatrix}$
$= \widehat i (0.6) - \widehat (0.4) + \widehat k (-0.9)$
$= 0.6 \widehat i - 0.4 \widehat j - 0.9 \widehat k$
$U =- \vec P \cdot \vec E$
$=- (2 \widehat i + 3 \widehat j) \cdot (0.3 \widehat i + 0.2 \widehat k)$
$=- 0.6 J$
Consider,
the dipole rotated by 180$^o$.

The magnitude of dipole moment will not change only its direction will change.
$\therefore U=-\vec{P}.\vec{E}=|P||E|\ Sin\theta$
the max value of U is $|P||E|$
$=\sqrt{13} \times 10^{-6} \times \sqrt{13} \times 10^5
= 1.3 J.$
This is the maximum potential energy of the dipole.

Multiple choice physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

 An electric dipole of moment $P$ is placed in the position of stable equilibrium in uniform electric field of intensity $E$. It is rotated through an angle $\theta$ from the initial position. The potential energy of electric dipole in the position is

  1. $\mathrm { pE } \cos \theta$
  2. $\mathrm { pE } \sin \theta$
  3. $\mathrm { pE } ( 1 - \cos \theta )$
  4. $\mathrm {- pE } \cos \theta$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Stable equilibrium is at theta = 0. Potential energy U = -pE cos(theta). Rotating to angle theta gives U(theta) = -pE cos(theta). The change in potential energy relative to the stable position is U(theta) - U(0) = -pE cos(theta) - (-pE) = pE(1 - cos(theta)).

Multiple choice physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

 A small dipole is placed is located at the center of an imaginary spherical Gaussian surface (radius R) with its dipole moment in +X-direction . Let $E _{max}$ & $E _{min}$  be maximum & maximum possible magnitude of field over the surface. 
Statement 1:   Number of points where E = $E _{max}$ is infinite.
Statement 2:    Number of points where E = $E _{min}$ is two.

  1. Both 1 and 2 are correct

  2. Both 1 and 2 are incorrect

  3. Only 1 is correct

  4. Only 2 is correct
Reveal answer Fill a bubble to check yourself
C Correct answer