Question

Two concentric spherical surfaces \(P_1\) and \(P_2\) enclose charges \(\dfrac{Q}{2}\) and \(4Q\) as shown in the figure. If \(\phi_1\) and \(\phi_2\) are the electric fluxes linked with the surfaces \(P_1\) and \(P_2\) respectively, then

• A. \(\phi_2=9\phi_1\)
• B. \(\phi_1=9\phi_2\)
• C. \(\phi_2=2\phi_1\)
• D. \(\phi_1=2\phi_2\)

Hint:

By Gauss's law, \[ \phi=\frac{q_{\text{enclosed}}}{\varepsilon_0}. \] Electric flux through a closed surface depends only on the total charge enclosed by that surface, not on the radius or exact position of the charge inside.

Answer 1

The Correct Option is A

Step 1: Use Gauss's law.
According to Gauss's law, the electric flux through a closed surface is \[ \phi=\frac{q_{\text{enclosed}}}{\varepsilon_0} \] where \(q_{\text{enclosed}}\) is the total charge enclosed by that closed surface.

Step 2: Find the flux through surface \(P_1\).
From the figure, the inner spherical surface \(P_1\) encloses only the charge \[ \frac{Q}{2}. \] Therefore, \[ \phi_1=\frac{Q/2}{\varepsilon_0} \] \[ \phi_1=\frac{Q}{2\varepsilon_0} \]

Step 3: Find the flux through surface \(P_2\).
The outer spherical surface \(P_2\) encloses both charges: \[ \frac{Q}{2} \] and \[ 4Q. \] So the total charge enclosed by \(P_2\) is \[ q_{\text{enclosed}}=\frac{Q}{2}+4Q \] \[ q_{\text{enclosed}}=\frac{Q}{2}+\frac{8Q}{2} \] \[ q_{\text{enclosed}}=\frac{9Q}{2} \] Therefore, \[ \phi_2=\frac{9Q/2}{\varepsilon_0} \] \[ \phi_2=\frac{9Q}{2\varepsilon_0} \]

Step 4: Compare \(\phi_1\) and \(\phi_2\).
Since \[ \phi_1=\frac{Q}{2\varepsilon_0} \] and \[ \phi_2=\frac{9Q}{2\varepsilon_0}, \] we get \[ \phi_2=9\phi_1 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\phi_2=9\phi_1} \]