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

An electric dipole consists of two opposite charges each of magnitude $2\mu C$ separated by a distance $1cm$. The dipole is placed in an external field of $10^3N/C$. The maximum torque on the dipole is

  1. $1\times 10^{-5}N-m$
  2. $2\times 10^{-5}N-m$
  3. $0.5\times 10^{-5}N-m$
  4. $Zero$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Torque tau = pE sin(theta). Max torque occurs at theta = 90 degrees, so tau_max = pE. p = q * d = (2*10^-6 C) * (0.01 m) = 2*10^-8 C-m. E = 10^3 N/C. tau_max = (2*10^-8) * (10^3) = 2*10^-5 N-m.

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

The relation connecting the energy U and distance r between dipole and induced dipole is :

  1. $U\propto r$
  2. $U\propto r^{2}$
  3. $U\propto r^{-6}$
  4. $U\propto r^{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The potential energy for the dipole-dipole interaction is given by $\displaystyle U=-\dfrac{2p _1^2p _2^2}{3(4\pi\epsilon _0)^2k _BT r^6}$
thus, $U \propto r^{-6}$

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 moment $ \overrightarrow { P }  $ is lying a uniform electric field $ \overrightarrow { E }  $ .The work done in rotation the dipole by $ 37^o $

  1. $ \dfrac {2}{5} PE $
  2. $ - \dfrac {2}{5} PE $
  3. $ \dfrac {PE}{5} $
  4. $ \dfrac {3}{5} PE $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Work done W = pE(cos(theta1) - cos(theta2)). Assuming rotation from 0 to 37 degrees, W = pE(cos(0) - cos(37)) = pE(1 - 4/5) = pE/5.

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 is placed in an electric field generated by a point charge then

  1. Then net electric force on the dipole must be zero

  2. The net electric force on the dipole may be zero

  3. The torque on the dipole due to the field may be zero

  4. Both (2) and (3)
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a non-uniform field (like that of a point charge), the net force on a dipole is generally non-zero. However, the torque can be zero if the dipole is aligned with the radial field line.

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 when placed in a uniform electric field $E$ will have a minimum potential energy if the dipole moment makes the following angle with $E$

  1. $\pi$
  2. $\pi /2$

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

${ U } _{ p }=-p\bullet E=-pE\cos { \theta  } $
${ \left( { U } _{ p } \right)  } _{ minimum }=-pE$
$\theta ={ 0 }^{ o }$${ U } _{ p }=-p\bullet E=-pE\cos { \theta  } $
${ \left( { U } _{ p } \right)  } _{ minimum }=-pE$
$\theta ={ 0 }^{ o }$

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 a 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 p. E

  2. zero and minimum

  3. q. E and maximum

  4. 2q. E and minimum
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$F = p\dfrac{dE} {dr} = 0 \left ( \because E = constant \right )$
$u = -\overrightarrow{p} \overrightarrow{E} = -PE \left ( minimum \right )$

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

Intensity of an electric field (E) depends on distance $r$. In case of dipole, it is related as :

  1. $ E \propto \cfrac{1}{r}$
  2. $ E \propto \cfrac{1}{r^{2}}$
  3. $ E \propto \cfrac{1}{r^{3}}$
  4. $ E \propto \cfrac{1}{r^{4}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Intensity of electric field due to a Dipole
$ E = \cfrac{p}{4\pi \varepsilon _{0}r^{3}} \sqrt{3cos^{2 }\theta+1}\Rightarrow E \propto \cfrac{1}{r^{3}}$

So, we can just dimensionally tell that Electric field will be inversely proportional to third power of $r$.

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

A point charge $Q$ lies on the perpendicular bisector of an electric dipole of dipole $p$. If the distance of $Q$ from the dipole is $r$ (much larger than the size of the dipole).then the electric field at $\theta$ is proportional to :

  1. $P^{2}$ and $r^{-3}$
  2. $P$ and $r^{-2}$
  3. $P^{-1}$ and $r^{-2}$
  4. $P$ and $r^{-3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l} As\, \, we\, \, have, \ if\, \, r>1 \ { P _{ axi } }=\frac { 1 }{ { 4\pi { E _{ 0 } } } } \frac { { 2P } }{ { { r^{ 3 } } } }  \ { V _{ axi } }=\frac { 1 }{ { 4\pi { E _{ 0 } } } } \frac { P }{ { { r^{ 2 } } } }  \ Where\, \, in, \ Angle\, \, between\, \, { P _{ axi } }\, \, and\, \, P\, \, is\, 0. \ { E _{ equatorial } }=\frac { { kp } }{ { { r^{ 3 } } } }  \ i.e\, \, \, E\propto p \ and\, \, P\propto { r^{ -3 } } \end{array}$

Hence, Option $D$ is correct answer.

Multiple choice capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

Two points A and B lying on Y- axis at distances 12.3 cm and 12.5 cm from the origin. The potentials at these points are 56V and 54.8V respectively, then the component of force on a charge of $4\mu C$ placed at A along Y- axis will be

  1. 0.12 N
  2. 48$*{10^{ - 3}}$ N
  3. $24*{10^{ - 4}}N$
  4. $96*{10^{ - 2}}N$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electric field E = -dV/dy = -(54.8 - 56) / (12.5 - 12.3) = -(-1.2) / 0.2 = 6 V/cm = 600 V/m. Force F = qE = 4 * 10^-6 * 600 = 24 * 10^-4 N.

Multiple choice capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

Inside a hollow charged spherical conductor, the electric field is found to be.

  1. Proportional to the distance from the centre

  2. A function of the area of the sphere

  3. Zero

  4. A function of the charge density of the sphere
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Gauss's Law, the electric field inside a hollow charged conductor is zero because there is no enclosed charge.

Multiple choice physics static electricity exchange of electrons methods used for charging electric charge

Two identical conducting balls having positive charges $q _1$ and $q _2$ are separated by a distance r. If they are made to touch each other and then separated to the same distance, the force between them will be

  1. less than before

  2. same as before

  3. more than before

  4. zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If they are made to touch each other, the electrostatic conduction will occur. As a result the total charge will get distributed equally on both the balls because both the balls are identical in nature. 


Initially, $F \propto q _1q _2$      (Since $F = \dfrac{kq _1q _2}{r^2}$)


Now after making contact, charge on each ball will become $Q _1=Q _2=\dfrac{q _1+q _2}{2}$
Now, $F' \propto (\dfrac{q _1 + q _2}{2})^2$
(Because they are separated to the same distance) 


From above we can say that $F' > F$

Multiple choice physics static electricity exchange of electrons methods used for charging electric charge

 A conducting wire is connected between two conducting spheres of equal size have a charge of -3C and +1C respectively. Find out the new charge on each sphere ? 

  1. -4C

  2. +4C

  3. -1C

  4. +1C

  5. Zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When the two spheres are connected by a wire, the charge will flow between them till the potential between them will be same. Thus, the charge on each sphere is the average charge i.e $Q _{av}=\dfrac{-3+1}{2}=-1 C$

Multiple choice physics static electricity exchange of electrons methods used for charging electric charge

A solid sphere of radius R has a charge Q distributed in its volume with a charge density, $\rho ={ kr }^{ a }$, where k and a are constants and r is the distance from its centre. If the electric field at $r=\dfrac { R }{ 2 } $ is $\dfrac { 1 }{ 8 } $ times that at r=R, then the value of a is 

  1. 2

  2. 4

  3. 6

  4. 7
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Gauss's law, the enclosed charge q(r) for a sphere with charge density rho = k * r^a is found by integrating the charge density over the volume. The electric field E(r) is proportional to q(r) / r^2. Setting up the ratio E(R/2) / E(R) = (1/8) and solving for a yields a = 2.

Multiple choice physics static electricity exchange of electrons methods used for charging electric charge

The linear charge density of a thin metallic rod varies with the distance $'x'$ from one end as $\lambda  = {\lambda _0}{x^2}\left( {0 \leqslant x \leqslant l} \right).$ The total charge on the rod is:

  1. $\dfrac{{{\lambda _0}{l^3}}}{3}$
  2. $\dfrac{{{\lambda _0}{l^4}}}{3}$
  3. $\dfrac{{2{\lambda _0}{l^3}}}{3}$
  4. $\dfrac{{{\lambda _0}{l}}}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\lambda =\dfrac {\lambda _0 x}{L}$
$Q=\displaystyle \int _0^L \lambda \ dx$
$=\displaystyle \int _0^L \dfrac {\lambda _0 x}{L}dx$
$=\dfrac {\lambda _0}{L} \dfrac {x^2}{2}\displaystyle \int _0^L$
$=\dfrac {\lambda _0}{L}\times \dfrac {L^2}{2}$
$=\dfrac {\lambda _0L}{2}$