Question

Two charges of magnitudes -5Q and +3Q are located at points (2a, 0) and (5a, 0) respectively. What is the electric flux due to these charges through a sphere of radius 4a with its centre at the origin?

Answer 1

The electric flux due to these charges through the sphere is $\frac{-5Q}{\epsilon_0}$.

Answer 2

• Concept: Gauss's Law states that the total electric flux ($\Phi$) through any closed surface is equal to the total electric charge ($Q_{enclosed}$) enclosed within that surface divided by the permittivity of free space ($\epsilon_0$).

  $\Phi = \frac{Q_{enclosed}}{\epsilon_0}$

• Identify the closed surface: A sphere of radius $R = 4a$ with its center at the origin $(0,0)$.

• Locate the charges relative to the sphere:

  1. Charge $q_1 = -5Q$ is located at $(2a, 0)$. The distance of $q_1$ from the origin is $\sqrt{(2a)^2 + 0^2} = 2a$. Since $2a < 4a$ (the radius of the sphere), the charge $q_1$ is inside the sphere.
  2. Charge $q_2 = +3Q$ is located at $(5a, 0)$. The distance of $q_2$ from the origin is $\sqrt{(5a)^2 + 0^2} = 5a$. Since $5a > 4a$ (the radius of the sphere), the charge $q_2$ is outside the sphere.

• Calculate the total enclosed charge ($Q_{enclosed}$): According to Gauss's Law, only the charges enclosed within the surface contribute to the net flux. Therefore, $Q_{enclosed} = q_1 = -5Q$.

• Apply Gauss's Law: $\Phi = \frac{Q_{enclosed}}{\epsilon_0} = \frac{-5Q}{\epsilon_0}$

The electric flux due to these charges through the sphere is $\frac{-5Q}{\epsilon_0}$.

Explanation of the solution:

The electric flux through a closed surface is determined solely by the net charge enclosed within that surface, as per Gauss's Law ($\Phi = Q_{enclosed}/\epsilon_0$). We identify the positions of the two charges relative to the given sphere. The charge $-5Q$ at $(2a, 0)$ is inside the sphere of radius $4a$ centered at the origin because its distance from the origin ($2a$) is less than the radius. The charge $+3Q$ at $(5a, 0)$ is outside the sphere because its distance from the origin ($5a$) is greater than the radius. Therefore, only the $-5Q$ charge contributes to the net flux through the sphere.