Question

Two charges of magnitude -2 q and + 5 q are located at points a,0 and 4a,0 respectively.What is the electric flux due to these charges

Answer 1

Answer: The electric flux is $\phi = \dfrac{3q}{\epsilon_{o}}$

Explanation: According to Gauss's law , the total flux of an electric field through a closed surface is proportional to the total electric charge.

Given that,

First charge of magnitude = -2q

Second charge of magnitude = +5q

Located at points = (a,0), (4a,0)

According to Gauss's law

$\phi = \oint \vec{E}\cdot \vec{ds}$

$\oint \vec{E}\cdot \vec{ds} = \dfrac{Q}{\epsilon_{o}}$....(I)

The total charge is

$Q = -2q + 5q$

$Q = 3q$

put the value of Q in equation (I)

$\phi = \dfrac{3q}{\epsilon_{o}}$

Hence, the electric flux is $\phi = \dfrac{3q}{\epsilon_{o}}$.