Conducting Sphere and Dipole Problem

[jojosg]

Homework Statement

Help with conducting sphere with dielectrics & induced charges

Relevant Equations

V = (1/(4πε₀)) * (q/r)

Hi, I'm stuck at this question, please help.

Attempt to the Conducting Sphere and Dipole Problem

(a) Electric Field and Potential at O due to Induced Charges

$$V_O = 0$$

This potential is the sum of the potentials due to the real charges ($+q, -q$) and the induced charges on the sphere.

$$V_O = V_{\text{real}} + V_{\text{induced}} = 0$$

- Electric Field at O, $\vec{E}_O$: Since point O is inside a conductor in electrostatic equilibrium, the electric field there must be zero.

$$\vec{E}_O = 0$$

This field is also the sum of the fields from the real charges and the induced charges.

$$\vec{E}_O = \vec{E}_{\text{real}} + \vec{E}_{\text{induced}} = 0$$

(b) Induced Charge $q'$ after Removing $-q$ and Grounding the Sphere

$$\left(\frac {q'} {4\pi\epsilon_{\text{0}}a}\right)+ \left(\frac {q} {4\pi\epsilon_{\text{0}}R}\right) = 0$$

Solving for $q'$:

$$\boxed{q' = -q\cdot\left( \frac R a \right)}$$

 

[TSny]

Homework Statement: Help with conducting sphere with dielectrics & induced charges
Relevant Equations: V = (1/(4πε₀)) * (q/r)

Your homework statement mentions dielectrics. But the actual problem statement does not mention dielectrics.

Hi, I'm stuck at this question, please help.

It's not clear to me where, specifically, you are stuck.

(a) Electric Field and Potential at O due to Induced Charges

$$V_O = 0$$

This potential is the sum of the potentials due to the real charges ($+q, -q$) and the induced charges on the sphere.

$$V_O = V_{\text{real}} + V_{\text{induced}} = 0$$

What is your reasoning for claiming that the total potential at point $O$ is zero?

The problem statement asks you to find the potential at $O$ due to the induced charge only.

- Electric Field at O, $\vec{E}_O$: Since point O is inside a conductor in electrostatic equilibrium, the electric field there must be zero.

$$\vec{E}_O = 0$$

This field is also the sum of the fields from the real charges and the induced charges.

$$\vec{E}_O = \vec{E}_{\text{real}} + \vec{E}_{\text{induced}} = 0$$

Yes, that looks good. But you still need to determine the field at $O$ due to the induced charge only (and include your reasoning).

(b) Induced Charge $q'$ after Removing $-q$ and Grounding the Sphere

$$\left(\frac {q'} {4\pi\epsilon_{\text{0}}a}\right)+ \left(\frac {q} {4\pi\epsilon_{\text{0}}R}\right) = 0$$

You should include some verbal statements that justify this equation.
Did you accidentally switch $q$ and $q'$ in this equation?

Solving for $q'$:

$$\boxed{q' = -q\cdot\left( \frac R a \right)}$$

OK. This is what you get if you switch $q$ and $q'$ in your previous equation.