Question

As shown in the figure, a point charge $Q$ is placed at the centre of conducting spherical shell of inner radius $a$ and outer radius $b$. The electric field due to charge $Q$ in three different regions I, II and III is given by: (I: r $<$ a, II: $a $<$ r $<$ b$, III: $r$>$b)$

• A. $E_I=0, E_{I I}=0, E_{I I I}=0$
• B. $E_I \neq 0, E_{I I}=0, E_{I I}=0$
• C. $E_I=0, E_{I I}=0, E_{I I I} \neq 0$
• D. $E_I \neq 0, E_{I I}=0, E_{I I I} \neq 0$

Answer 1

The Correct Option is D

Understanding the Setup:

1. A point charge \( Q \) is placed at the center of a conducting spherical shell with inner radius \( a \) and outer radius \( b \). We need to determine the electric field \( E \) in three different regions:
   • Region I: \( r < a \)
   • Region II: \( a < r < b \)
   • Region III: \( r > b \)

Region I (\( r < a \)):

1. - According to Gauss's Law, the electric field \( E \) inside a conductor is determined by the enclosed charge. Here, the charge enclosed by a Gaussian surface within radius \( r < a \) is \( Q \). - Therefore, \( E \neq 0 \) inside this region. - The electric field is given by: \(E = \frac{Q}{4\pi\varepsilon_0r^2}\)

Region II (\( a < r < b \)):

1. - In this region, we are inside the conducting shell but outside the center. - The nature of conductors is such that they will rearrange their charges to cancel out the internal electric field. - Therefore, \( E = 0 \) in this region.

Region III (\( r > b \)):

1. - For \( r > b \), the shell can be treated as a point charge at the center because of the spherical symmetry. - The charge \( Q \) effectively influences this region. - The electric field is given by: \(E = \frac{Q}{4\pi\varepsilon_0r^2}\) - Thus, \( E \neq 0 \) here.

Conclusion:

• The correct answer is \( E_I \neq 0 \), \( E_{II} = 0 \), \( E_{III} \neq 0 \).