Question

The electric flux from a cube of edge l is ϕ.What will be its value if edge of cube is made 2l and charge enclosed is halved-
A)4ϕ B)2ϕ C) ϕ D) ϕ/2

Answer 1

Answer on Question #41186, Physics, Electromagnetism

The electric flux from a cube of edge $l$ is $\varphi$. What will be its value if edge of cube is made 2l and charge enclosed is halved-

A) $4\varphi$

B) $2\varphi$

C) $\varphi$

D) $\varphi/2$

Solution

According Gauss's law the net flux from a cube is

$$\phi = \frac{Q}{\epsilon_0},$$

where $Q$ is a charge enclosed by a cube, $\epsilon_0$ is the vacuum permittivity.

If edge of cube is made 2l and charge enclosed is halved the net flux from a cube is

$$\phi_1 = \frac{\frac{Q}{2}}{\epsilon_0} = \frac{1}{2} \frac{Q}{\epsilon_0} = \frac{1}{2} \phi.$$

Notice that the edge length of the cube did not enter into the calculation.

Answer: D) $\frac{1}{2}\phi$

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