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A solid spherical conductor of radius R ...

A solid spherical conductor of radius R has a spherical cavity of radius at its centre. A charge is kept at the centre of the cavity. The charge at the inner surface, outer surface are respectively.

A

`+Q,-Q`

B

`-Q,+Q`

C

`Q, - Q`

D

`+Q,0`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the situation of a solid spherical conductor with a spherical cavity at its center, where a charge \( +q \) is placed at the center of the cavity. We will determine the charge on the inner and outer surfaces of the conductor. ### Step-by-Step Solution: 1. **Understanding the Setup**: - We have a solid spherical conductor with radius \( R \). - There is a spherical cavity of radius \( r \) at the center of this conductor. - A charge \( +q \) is placed at the center of the cavity. 2. **Electric Field Inside the Conductor**: - In electrostatics, the electric field inside a conductor in electrostatic equilibrium is zero. This means that there cannot be any electric field present in the material of the conductor itself. 3. **Induced Charge on the Inner Surface**: - Since there is a charge \( +q \) at the center, it will induce a charge on the inner surface of the cavity. - To neutralize the electric field due to the charge \( +q \) within the conductor, an equal and opposite charge must be induced on the inner surface of the cavity. - Therefore, the charge induced on the inner surface of the cavity will be \( -q \). 4. **Total Charge on the Conductor**: - The conductor as a whole must remain electrically neutral. If we denote the total charge of the conductor as \( Q \), we have: \[ Q = Q_{\text{inner}} + Q_{\text{outer}} \] - Since the inner surface has a charge of \( -q \), we can express this as: \[ Q = -q + Q_{\text{outer}} \] - For the conductor to be neutral, if \( Q \) is zero (assuming the conductor was initially uncharged), then: \[ 0 = -q + Q_{\text{outer}} \] - This leads to: \[ Q_{\text{outer}} = +q \] 5. **Final Charges**: - The charge on the inner surface of the cavity is \( -q \). - The charge on the outer surface of the conductor is \( +q \). ### Conclusion: - The charge on the inner surface is \( -q \). - The charge on the outer surface is \( +q \).
To solve the problem, we need to analyze the situation of a solid spherical conductor with a spherical cavity at its center, where a charge \( +q \) is placed at the center of the cavity. We will determine the charge on the inner and outer surfaces of the conductor. ### Step-by-Step Solution: 1. **Understanding the Setup**: - We have a solid spherical conductor with radius \( R \). - There is a spherical cavity of radius \( r \) at the center of this conductor. - A charge \( +q \) is placed at the center of the cavity. ...
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