Q: A spherical conductor of radius 10 cm has a charge of 3.2 \times 10^{-7} C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? (\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 N·m²/C²)

1.28 \times 10^5 N/C. Outside the sphere it acts as a point charge: E = \frac{1}{4\pi\varepsilon_0}\frac{q}{x^2} = \frac{9 \times 10^9 \times 3.2 \times 10^{-7}}{(0.15)^2} = 1.28 \times 10^5 N/C

Q: A pendulum bob of mass 30.7 \times 10^{-6} kg and carrying a charge 2 \times 10^{-8} C is at rest in a horizontal uniform electric field of 20000 V/m. The tension in the thread of the pendulum is (g = 9.8 m/s²)

5 \times 10^{-4} N. Weight mg = 30.7 \times 10^{-6} \times 9.8 = 3 \times 10^{-4} N; electric force qE = 2 \times 10^{-8} \times 20000 = 4 \times 10^{-4} N (horizontal). T = \sqrt{(mg)^2 + (qE)^2} = \sqrt{(3)^2 + (4)^2} \times 10^{-4} = 5 \times 10^{-4} N

Question 1

A dipole (+q on the left, −q on the right) is placed in a non-uniform electric field whose field lines converge towards the right, so the field is stronger at the +q end than at the −q end (|E_1| > |E_2|). In which direction will it move?

Accepted Answer

Towards the right as its potential energy will decrease. The field is stronger at the +q charge, so the net force on the dipole is towards the right. A system always moves so as to decrease its potential energy — hence it moves right and its PE decreases.

Question 2

A spherical conductor of radius 10 cm has a charge of 3.2 	imes 10^{-7} C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? (\frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9 N·m²/C²)

Accepted Answer

1.28 	imes 10^5 N/C. Outside the sphere it acts as a point charge:

E = \frac{1}{4\pi\varepsilon_0}\frac{q}{x^2} = \frac{9 	imes 10^9 	imes 3.2 	imes 10^{-7}}{(0.15)^2} = 1.28 	imes 10^5 N/C

Question 3

An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is

Accepted Answer

smaller. With a = \frac{qE}{m} and h = \frac{1}{2}at^2, the time t = \sqrt{\frac{2hm}{qE}} \propto \sqrt{m}.

Since m_p \gg m_e, the electron (smaller mass) takes a smaller time to fall.

Question 4

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 5

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 6

A pendulum bob of mass 30.7 	imes 10^{-6} kg and carrying a charge 2 	imes 10^{-8} C is at rest in a horizontal uniform electric field of 20000 V/m. The tension in the thread of the pendulum is (g = 9.8 m/s²)

Accepted Answer

5 	imes 10^{-4} N. Weight mg = 30.7 	imes 10^{-6} 	imes 9.8 = 3 	imes 10^{-4} N; electric force qE = 2 	imes 10^{-8} 	imes 20000 = 4 	imes 10^{-4} N (horizontal).

T = \sqrt{(mg)^2 + (qE)^2} = \sqrt{(3)^2 + (4)^2} 	imes 10^{-4} = 5 	imes 10^{-4} N

Electric and Field Lines — NEET Physics PYQ MCQs with Solutions

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