Question

The charge enclosed in the two cubes A and B is q. If ratio of the surface area of cubes A to cube B is 1 : 2, then the ratio of the flux associated with cube A to cube B will be:
1. 1 : 2
2. 2 : 1
3. 1 : 1
4. None of these

Answer 1

Correct Answer - Option 3 : 1 : 1

Concept:

Gauss's law:

According to Gauss law, the total electric flux linked with a closed surface called Gaussian surface is \(\frac{1}{ϵ_o}\) the charge enclosed by the closed surface.

\(\Rightarrow ϕ=\frac{Q}{ϵ_o}\)

Where ϕ = electric flux linked with a closed surface, Q = total charge enclosed in the surface, and ϵo = permittivity

Important points:

1. Gauss’s law is true for any closed surface, no matter what its shape or size.
2. The charges may be located anywhere inside the surface.

 

Calculation:

Given Q_A = Q_B = q, and A_A : A_B = 1 : 2

Where AA and AB = surface area of the cube A and the cube B respectively

By the Gauss law, if the total charge enclosed in a closed surface is Q, then the total electric flux associated with it will be given as,

\(\Rightarrow ϕ=\frac{Q}{ϵ_o}\)     -----(1)

By equation 1 it is clear that the total flux linked with the closed surface in which a certain amount of charge is placed does not depend on the shape and size of the surface.

By equation 1 the total flux associated with cube A is given as,

\(\Rightarrow ϕ_A=\frac{Q_A}{ϵ_o}\)

\(\Rightarrow ϕ_A=\frac{q}{ϵ_o}\)     -----(2)

By equation 1 the total flux associated with cube B is given as,

\(\Rightarrow ϕ_B=\frac{Q_B}{ϵ_o}\)

\(\Rightarrow ϕ_B=\frac{q}{ϵ_o}\)     -----(3)

By equation 2 and equation 3,

\(\Rightarrow \frac{ϕ_A}{ϕ_B}=\frac{q}{ϵ_o}\times\frac{ϵ_o}{q}\)

\(\Rightarrow \frac{ϕ_A}{ϕ_B}=\frac{1}{1}\)

Hence, option 3 is correct.