MCQs on Electrostatics: Coulomb’s Law & Gauss Law

Ans. d) \(\frac{Q}{6 \varepsilon_o}\)  x 10^-6

Explanation: According to Gauss’s law, 

ϕ = Q/ϵ₀

Where,

Q = total charge within the given surface,

ε₀ = the electric constant.

Therefore, the total flux is;

Φ = \(\frac{Q}{\varepsilon_o}\) x 10^-6.

As the cube has six sides, the electric flux from one side of the cube = \(\frac{Q}{\varepsilon_o}\)  x 10^-6 x \(\frac{1}{6}\) 

Φ = \(\frac{Q}{6 \varepsilon_o}\)  x 10^-6

Ans. b) Coulomb’s law.

Explanation: According to Coulomb’s law the magnitude of the force of attraction or repulsion between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges. Like charges repel and opposite charges attract each other.

Thus, Coulomb’s law gives us the force between two charged particles in electrostatics.

Ans. a) It increases by four times

Explanation: We know that, according to Coulomb’s law, the force of attraction or repulsion between two point charges is inversely proportional to the square of the distance between the charges.

F∝ \(\frac{1}{r^2}\)

F'=k \(\frac{q_1 q_2}{r_2}\)

As the distance is halved from its initial value:

⇒ F=k \(\frac{q_1q_2}{(\frac{r}{2})^2}\)

⇒ F= k \(\frac{q_1 q_2}{\frac{r}{4}}\)

⇒ F= k \(\frac{q_1 q_2 \times 4}{r^2}\)

Therefore, F= 4 F’

Ans. b) B the electric potential is maximum at point B.

Explanation: Electric field intensity and electric potential are related to each other by-

E=−dV/dx.....(i)

This equation shows the rate of change of potential with respect to the distance in the direction of the intensity. The negative sign states that the potential decreases with the distance measured in the direction of the electric field E.

So we can infer that the potential at point B is greater than that of points A and C.

Ans: (b) 1.6 x 10^-19 Coulomb

Explanation: The minimum charge on a particle that could possibly be is the charge of an electron. Any particle can have the least possible charge equal to the charge of an electron. Therefore, the charge of one electron is 1.6 x 10^-19 Coulomb which is also the minimum charge that a particle can have. 

Ans. a) The separation between the plates.

Explanation: The capacity of a parallel plate capacitor depends upon the following three factors:

1. Is inversely proportional to the separation between the plates. (1C∝1d.
2. Is directly proportional to the area of the plates. C∝A.
3. Is directly proportional to the dielectric constant of the medium C∝k.

Ans. d) V(C₂/(C₁ + C₂) 

Explanation: For two parallel plate capacitors connected in series, the total capacity is given by:-

C = \(\frac{1}{C_1} + \frac{1}{C_2}\)

C= \(\frac{C_2 + C_1}{C_1 C_2}\)

Then, Q = CV = ( \(\frac{C_2 + C_1}{C_1 C_2}\) ) V

Thus, the potential across C₁ = \(\frac{C_1 C_2 V}{C_1 + C_2} \times \frac{1}{C_1}\)

= \(\frac{C_2 V}{C_1 + C_2} \)

Ans. a) It experiences a net torque only.

Explanation: For an electric dipole kept in a uniform electric field, it experiences a net torque, according to the following equation:

τ=pEsin θ

=Torque on the dipole

P= Dipole moment of the dipole

E= Electric field intensity

θ = The angle made by the centre of the dipole from the direction of the electric field.

No net force is experienced by the dipole because the forces arising due to the equal and opposite charges of the dipole cancel out each other. Thus, the net translatory force is zero.

Ans. a) 4 × 10^-4 N

Explanation: The force between two infinite conductors carrying charge is given by:

 \(\frac{ \mu _ ° I_1 I_2 I}{ 2 \pi d}\)

2 × 10^-7 \(\frac{ I_1 I_2 I}{ 2 \pi d}\)

• I₁ , I₂ = Current carried by the two conductors
• l= Length 
• d= Distance between the conductors

For force per unit length :

\(\frac{F}{L}\)= \(\frac{ \mu _ ° I_1 I_2 I}{ 2 \pi d}\) = 2× 10^-7 \(\frac{20 ×20}{0.2}\) = 4× 10^-4 N m -. 

Ans. d) Scalar vector

Explanation: The dot product of two vectors is always a scalar quantity. While the cross product of a vector and a scalar is a vector quantity.

Here, electric flux is a dot product of electric field intensity and the area of the surface which is considered.

Φ = E • A 

So it is a scalar value.

The electric flux density is a vector quantity because it is a product of a scalar

And a vector E (Electric field intensity).

D = ε × E.