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A solid spherical conductor of radius R ...

A solid spherical conductor of radius R has a spherical cavity of radius a (a < R) at its centre. A charge +Q is kept at the center. The charge at the inner surface, outer and at a position `r (a lt r lt R)` are respectively

A

`+Q , -Q , 0`

B

`-Q , +Q , 0`

C

`0,Q,0`

D

`+Q,0,0`

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the situation step by step, considering the properties of conductors and electrostatics. ### Step 1: Understanding the Setup We have a solid spherical conductor with radius \( R \) that has a spherical cavity of radius \( a \) (where \( a < R \)) at its center. A charge \( +Q \) is placed at the center of this cavity. **Hint:** Remember that in electrostatics, charges redistribute themselves on conductors to maintain equilibrium. ### Step 2: Charge Induction on the Inner Surface When the charge \( +Q \) is placed at the center of the cavity, it induces a charge on the inner surface of the conductor. According to Gauss's law, the induced charge on the inner surface will be \( -Q \) to neutralize the effect of the charge \( +Q \) inside the cavity. **Hint:** The total charge inside a conductor must be zero when in electrostatic equilibrium. ### Step 3: Charge on the Outer Surface Since the conductor is neutral overall, the total charge on the conductor must remain zero. The charge \( -Q \) on the inner surface means that the outer surface must have a charge of \( +Q \) to balance it out. Thus, the charge on the outer surface of the conductor is \( +Q \). **Hint:** The outer surface charge is equal to the total charge placed inside the conductor plus any induced charges. ### Step 4: Charge at a Point \( r \) (where \( a < r < R \)) Now, we consider a point \( P \) located at a distance \( r \) from the center of the cavity, where \( a < r < R \). According to the properties of conductors, the electric field inside a conductor is zero. Therefore, there is no charge present at this point \( P \). **Hint:** Remember that inside a conductor, the electric field is zero, implying that the charge density is also zero. ### Summary of Charges 1. **Charge at the inner surface of the conductor**: \( -Q \) 2. **Charge at the outer surface of the conductor**: \( +Q \) 3. **Charge at a position \( r \) (where \( a < r < R \)**: \( 0 \) ### Final Answer - Charge at the inner surface: \( -Q \) - Charge at the outer surface: \( +Q \) - Charge at position \( r \): \( 0 \)
To solve the problem, we need to analyze the situation step by step, considering the properties of conductors and electrostatics. ### Step 1: Understanding the Setup We have a solid spherical conductor with radius \( R \) that has a spherical cavity of radius \( a \) (where \( a < R \)) at its center. A charge \( +Q \) is placed at the center of this cavity. **Hint:** Remember that in electrostatics, charges redistribute themselves on conductors to maintain equilibrium. ### Step 2: Charge Induction on the Inner Surface ...
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