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 applications of gauss's law coulomb's law physics

A thinwalled, spherical conducting shell S of radius R is given charge Q. The same amountof charge is also placed at its centre C. Which of the following statements are correct?

  1. On the outer surface of S, the charge density is $\displaystyle \frac {Q} {2 \pi R^2} $

  2. The electric field is zero at all points inside S

  3. At a point just outside S, the electric field is double the field at a point just inside S

  4. At any point inside S, the electric field is inversely proportional to the square of its distance from C
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

because of charge Q at center there will be induced charge -Q at inner surface of sphere. hence charge density $\dfrac{2Q}{4\pi r^2} = \dfrac{Q}{2\pi r^2}$

because of 2Q charge outside the electric field is double that of inside.
At any point inside S, the electric field is inversely proportional to the square of its distance from C

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

Two sphere's are isolated from each other. They each have an identical net positive charge and have the same radius, however, one sphere is solid and insulating, while the other is a hollow conducting sphere whose charge is uniformly distributed.
For which sphere is the electric field the greatest distance $x$ from the center of the spheres?
Assume $x$ is less than the radius of the spheres.

  1. The conducting hollow sphere has a greater E-field.

  2. The insulating solid sphere has a greater E field.

  3. Both spheres have the same E field.

  4. Neither sphere would cause there to be an Electric field.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The charge in a conducting hollow sphere resides on the surface only. Hence on applying Gauss Law on a spherical surface enclosed in the sphere, we get electric field to be zero inside it.

For a solid insulating sphere with uniform charge distribution, finite electric field exists in the sphere.
Hence correct answer is option B.

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

As one penetrates through uniformly charged conducting sphere, what happens to the electric field strength:

  1. decreases inversely as the square of the distance

  2. decreases inversely as the distance

  3. becomes zero

  4. increases inversely as the square of distance
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electric field strength inside the uniform charged sphere is zero.

$\therefore$  As one penetrates through uniformly charged sphere, electric field strength inside the sphere becomes zero. 

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 motional emf physics

An electric charge $+ q$ moves with velocity $\bar{V}=3\hat{i}+4\hat{j}+\hat{k}$, in an electromagnetic field given by $\bar{E}=3\hat{i}+\hat{j}+\hat2{k}$ $\bar{B}=\hat{i}+\hat{j}+\hat3{k}$.The y-component of the force experienced by +q is:

  1. $-7 q$
  2. $11 q$
  3. $5 q$
  4. $3 q$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

Velocity, $v=\left( 3\hat{i}+4\hat{j}+\hat{k} \right)$

Electric field, $E=3\hat{i}+\hat{j}+2\hat{k}$

Magnetic field, $B=\hat{i}+\hat{j}+3\hat{k}$

Force,

$ F=q\left( \text{E+v}\times \text{B} \right) $

$ F=q\left( 3\hat{i}+\hat{j}+2\hat{k}+\left( (3\hat{i}+4\hat{j}+\hat{k})\times (\hat{i}+\hat{j}+3\hat{k} \right) \right) $

$ F=q\left( 3\hat{i}+\hat{j}+2\hat{k}+11\hat{i}-8\hat{j}-\hat{k} \right) $

$ F=q\left( 14\hat{i}-7\hat{j}+\hat{k} \right) $

So, y-component of force experienced by +q is

${{F} _{y}}=-7q$

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 deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

Cathode rays enter an electric field normal to the field. Then their path in the electric field is :

  1. A parabola

  2. A circle

  3. A straight line

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

Electric field applies a constant force  $eE$ on the electrons (i.e. cathode rays) perpendicular to the beam of electrons due to which the electrons will trace a parabolic path.

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.