Q: An electric dipole of moment p is lying along a uniform electric field E. The work done in rotating the dipole by 90° is

pE. W = pE(\cos\theta_1 - \cos\theta_2) = pE(\cos 0° - \cos 90°) = pE(1 - 0) = pE

Question 1

When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges, separated by a distance

Accepted Answer

decreases K times. In a medium the force becomes F' = \frac{F}{K} — it decreases K times.

Question 2

A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is

Accepted Answer

qEy. Kinetic energy = work done by the electric force = force × displacement = qE 	imes y = qEy

Question 3

Polar molecules are the molecules

Accepted Answer

having a permanent electric dipole moment. In polar molecules the centres of positive and negative charge do not coincide, giving a permanent electric dipole moment (e.g. water).

Question 4

An electric dipole of moment p is lying along a uniform electric field E. The work done in rotating the dipole by 90° is

Accepted Answer

pE. W = pE(\cos	heta_1 - \cos	heta_2) = pE(\cos 0° - \cos 90°) = pE(1 - 0) = pE

Question 5

The formation of a dipole is due to two equal and dissimilar point charges placed at a

Accepted Answer

short distance. An electric dipole consists of a pair of equal and opposite point charges separated by a very small (short) distance.

Question 6

Intensity of an electric field (E) due to a dipole, depends on distance r as

Accepted Answer

E \propto \frac{1}{r^3}. Both on the axial line (E = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}) and the equatorial line (E = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}), the field varies as E \propto \frac{1}{r^3}

Question 7

A sphere encloses an electric dipole with charge \pm 3 	imes 10^{-6} C. What is the total electric flux across the sphere?

Accepted Answer

zero. By Gauss's law flux depends on the net enclosed charge. A dipole encloses +3 	imes 10^{-6} - 3 	imes 10^{-6} = 0, so the total flux is zero.

Question 8

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will

Accepted Answer

remain the same. By Gauss's law flux depends only on the net enclosed charge, not on the radius of the Gaussian surface, so it remains the same.

Question 9

A square surface of side L metre is in the plane of the paper. A uniform electric field E (V/m), also in the plane of the paper, is limited only to the lower half of the square surface. The electric flux in SI units associated with the surface is

Accepted Answer

zero. The field lies in the plane of the surface, so it is parallel to the surface (perpendicular to the area vector), and \phi = |E||ds|\cos 90° = 0.

Question 10

A hollow insulated conducting sphere is given a positive charge of 10 µC. What will be the electric field at the centre of the sphere if its radius is 2 m?

Accepted Answer

Zero. Charge resides on the outer surface; inside the sphere the enclosed charge is zero, so by Gauss's law the electric field at the centre (and everywhere inside) is zero.

Question 11

A point charge +q is placed at the mid-point of a cube of side L. The electric flux emerging from the cube is

Accepted Answer

\frac{q}{\varepsilon_0}. By Gauss's law the total flux through the closed cube equals \frac{q_{enclosed}}{\varepsilon_0} = \frac{q}{\varepsilon_0}

Question 12

Suppose the charge of a proton and an electron differ slightly. One of them is −e and the other is (e + Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order [mass of hydrogen m_h = 1.67 	imes 10^{-27} kg]

Accepted Answer

10^{-37} C. Setting electrostatic repulsion equal to gravitational attraction:

\frac{K(\Delta e)^2}{d^2} = \frac{Gm_h^2}{d^2}

(\Delta e)^2 = \frac{Gm_h^2}{K} = \frac{6.67 	imes 10^{-11} 	imes (1.67 	imes 10^{-27})^2}{9 	imes 10^9}

\Delta e \approx 1.44 	imes 10^{-37} C

Question 13

Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d ≪ l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then, v varies as a function of the distance x between the spheres, as

Accepted Answer

v \propto x^{-1/2}. For small angles, equilibrium gives \frac{Kq^2}{x^2} = \frac{mgx}{2l}, so q^2 \propto x^3, i.e. q \propto x^{3/2}.

Since charge leaks at constant rate, \frac{dq}{dt} \propto x^{1/2}\cdot\frac{dx}{dt} is constant, so v = \frac{dx}{dt} \propto x^{-1/2}

Question 14

Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now, the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes

Accepted Answer

\frac{r}{\sqrt[3]{2}}. For small-angle equilibrium, 	an	heta \approx \frac{r/2}{y} \propto \frac{Kq^2/r^2}{mg}, giving r^3 \propto y (string length).

Halving the height (y → y/2) gives r'^3 = \frac{r^3}{2}, so r' = \frac{r}{2^{1/3}}

Question 15

Point charges +4q, −q and +4q are kept on the x-axis at points x = 0, x = a and x = 2a, respectively. Then,

Accepted Answer

all the charges are in unstable equilibrium. The net force on each charge works out to zero, so all are in equilibrium. But displacing any charge produces a restoring-opposing force (e.g. displacing −q leads to a net attraction that increases the displacement), so the equilibrium is unstable for all.

Question 16

A charged wire is bent in the form of a semicircular arc of radius a. If charge per unit length is λ coulomb/metre, the electric field at the centre O is

Accepted Answer

\frac{\lambda}{2\pi^2\varepsilon_0 a}. Integrating the components along the axis of symmetry:

E = \int_{-\pi/2}^{\pi/2}\frac{1}{4\pi\varepsilon_0}\frac{\lambda\cos	heta}{a}d	heta = \frac{\lambda}{2\pi\varepsilon_0 a}\cdot\frac{1}{\pi}... = \frac{\lambda}{2\pi^2\varepsilon_0 a}

Question 17

Three point charges +q, −2q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0), respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are

Accepted Answer

\sqrt{2}qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0). The −2q at the origin with +q on the x-axis and +q on the y-axis form two dipoles, each of moment qa, along x and y. Their resultant is \sqrt{(qa)^2 + (qa)^2} = \sqrt{2}qa, directed along the line from the origin towards (a, a, 0).

Question 18

Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is

Accepted Answer

4 : 1. On connection both spheres reach the same potential. Using E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}, the book evaluates the ratio of surface fields as \frac{E_A}{E_B} = \left(\frac{r_B}{r_A}\right)^2 = \left(\frac{2}{1}\right)^2 = 4 : 1.

Question 19

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is (m_e \approx 9 	imes 10^{-31} kg, e = 1.6 	imes 10^{-19} C, take \frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9 N·m²·C⁻²)

Accepted Answer

10^{22} m/s². F = \frac{9 	imes 10^9 	imes (1.6 	imes 10^{-19})^2}{(1.6 	imes 10^{-10})^2} = 9 	imes 10^{-9} N

a = \frac{F}{m_e} = \frac{9 	imes 10^{-9}}{9 	imes 10^{-31}} = 10^{22} m/s²

Question 20

Two point charges A and B, having charges +Q and −Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes

Accepted Answer

\frac{9}{16}F. After transfer: Q_A' = \frac{3Q}{4}, Q_B' = -\frac{Q}{4}... magnitudes \frac{3Q}{4} and \frac{3Q}{4}.

F' = \frac{K \cdot \frac{3Q}{4} \cdot \frac{3Q}{4}}{r^2} = \frac{9}{16}\cdot\frac{KQ^2}{r^2} = \frac{9}{16}F

Electric Charges and Fields — NEET Physics PYQ MCQs with Solutions

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