Question

A small spherical shell \( S_1 \) has point charges \( q_1 = -3 \, \mu C \), \( q_2 = -2 \, \mu C \) and \( q_3 = 9 \, \mu C \) inside it. This shell is enclosed by another big spherical shell \( S_2 \). A point charge \( Q \) is placed in between the two surfaces \( S_1 \) and \( S_2 \). If the electric flux through the surface \( S_2 \) is four times the flux through surface \( S_1 \), find charge \( Q \).

Hint:

In Gauss’s law, the electric flux through a surface depends on the net charge enclosed within that surface.

Answer 1

Gauss's law states that electric flux through a surface is directly proportional to the net charge enclosed: \( \Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0} \). The total charge enclosed by \( S_2 \) comprises the charges within \( S_1 \) plus \( Q \), the charge situated between \( S_1 \) and \( S_2 \). Consequently, the total enclosed charge for \( S_2 \) is: \[ Q_{\text{enc}} = q_1 + q_2 + q_3 + Q \] The flux through \( S_1 \) is proportional to its enclosed charge: \[ \Phi_{S_1} = \frac{q_1 + q_2 + q_3}{\varepsilon_0} \] The flux through \( S_2 \) is four times that of \( S_1 \): \[ \Phi_{S_2} = 4 \cdot \Phi_{S_1} = \frac{4 \cdot (q_1 + q_2 + q_3)}{\varepsilon_0} \] Applying Gauss's law to \( S_2 \) yields: \[ \Phi_{S_2} = \frac{q_1 + q_2 + q_3 + Q}{\varepsilon_0} \] Equating the two expressions for \( \Phi_{S_2} \) gives: \[ \frac{4 \cdot (q_1 + q_2 + q_3)}{\varepsilon_0} = \frac{q_1 + q_2 + q_3 + Q}{\varepsilon_0} \] Solving for \( Q \): \[ 4 \cdot (q_1 + q_2 + q_3) = q_1 + q_2 + q_3 + Q \] \[ Q = 3 \cdot (q_1 + q_2 + q_3) \] Substituting the given charge values: \[ Q = 3 \cdot (-3 \, \mu C - 2 \, \mu C + 9 \, \mu C) = 3 \cdot 4 \, \mu C = 12 \, \mu C \] The value of charge \( Q \) is determined to be \( 12 \, \mu C \).