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: Apply Gauss's law for each surface.
φ = q_enclosed/ε₀.

Step 2: Flux through P₁ encloses Q/2.
φ₁ = (Q/2)/ε₀ = Q/(2ε₀).

Step 3: Flux through P₂ encloses Q/2 + 4Q = 9Q/2.
φ₂ = (9Q/2)/ε₀ = 9Q/(2ε₀).

Step 4: Compare fluxes.
φ₂ = 9φ₁.

Step 5: Final Answer:
φ₂ = 9φ₁.