AIIMS 2005 Physics Electric Field MCQ Question

To analyze the electric field produced by two concentric conducting spherical shells A and B, we need to apply Gauss's law and understand the behavior of electric fields in and around conductors.

Given:

• Radius of shell A: $r_a$
• Radius of shell B: $r_b$ (with $r_b > r_a$)
• Charge on shell A: $Q_a$
• Charge on shell B: $-Q_b$ (where $|Q_b| > |Q_a|$)

Step 1: Understanding the Electric Field in Different Regions

1. Inside shell A ($r < r_a$):

   • The electric field $E$ inside a conductor is zero. Therefore, for any point where $r < r_a$, we have:
   $$E = 0$$

2. Between the shells ($r_a < r < r_b$):

   • Here, we need to consider the effect of the charges on both shells. According to Gauss's Law, the electric field $E$ at a distance $r$ from the center (where $r_a < r < r_b$) is given by:
   $$E = \frac{Q_{\text{enc}}}{4\pi \epsilon_0 r^2}$$
   • The enclosed charge $Q_{\text{enc}}$ in this region is simply the charge on shell A, $Q_a$. Thus:
   $$E = \frac{Q_a}{4\pi \epsilon_0 r^2} \quad \text{for } r_a < r < r_b$$

3. Outside shell B ($r > r_b$):

   • For points outside both shells, the total enclosed charge is $Q_a + (-Q_b) = Q_a - Q_b$. Since $|Q_b| > |Q_a|$, we have $Q_a - Q_b < 0$:
   $$E = \frac{Q_a - Q_b}{4\pi \epsilon_0 r^2} \quad \text{for } r > r_b$$
   • This indicates that the electric field direction is towards the center of the shells because the net charge is negative.

Step 2: Analyzing the Electric Field Behavior

• At $r = r_a$: The electric field just outside shell A is positive, as derived.
• At $r = r_b$: The electric field just outside shell B becomes negative since $Q_a - Q_b < 0$.

Step 3: Summary of Electric Field Behavior

1. For $r < r_a$: $E = 0$
2. For $r_a < r < r_b$: $E$ is positive and decreases with $\frac{1}{r^2}$.
3. For $r > r_b$: $E$ becomes negative, also decreasing with $\frac{1}{r^2}$.

Conclusion

The electric field starts at zero, increases positively as we approach shell A, and then decreases to zero just before shell B. After shell B, the field becomes negative and decreases in magnitude as we move further away.

Correct Option

The correct graph representation for this behavior is Graph (c). It reflects:

• Zero electric field inside shell A.
• Positive electric field between the shells.
• A negative electric field outside shell B.

Incorrect Options

• Graph (a) and Graph (b): These graphs do not capture the transition from positive to negative correctly.
• Graph (d): This likely shows an incorrect continuous behavior without the transition from positive to negative, which does not match our analysis.

Thus, the correct answer is C (Graph (c)) because it accurately represents the behavior of the electric field in the described system.