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.

 

[jojosg]

Maybe just stuck at part a with induced charge. For part b I referred to this in my teachers powerpoint.

 

[jojosg]

(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.

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

Reasoning:

1. Conductor as an Equipotential: The hollow spherical shell B is a conductor. In electrostatic equilibrium, every point on and inside a conductor's material is at the same potential.
2. Potential of Shell B: This potential is constant throughout the shell's material.
3. Point O is Inside the Conductor: Point O, the center of the sphere, lies inside the conducting material of shell B (since R<R1, and the shell occupies the space from R1to R2).
4. Conclusion: Therefore, point O must be at the same potential as the shell itself.

$$V_O=V_B$$

Field from the real charge +q on sphere A:
By Gauss's law, a charge +q uniformly distributed on a shell of radius R produces a field at a point r>R as if it were a point charge at the center. Since O is at r=0, this formula doesn't apply directly. However, by symmetry, a uniform spherical shell produces zero electric field everywhere inside the shell, including at the center.

Net Field at O is Zero:

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

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

Reasoning: The field from the real charge +q is zero at O. For the net field at O to be zero, the field from the induced charges must also be zero. The induced charges arrange themselves on the spherical surfaces precisely to cancel any field inside the conductor, which includes point O.

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

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

Verbal Reasoning:

1. Initial State: A point charge +q is placed a distance a from the center Oof an isolated, uncharged spherical conductor of radius R. This induces a charge −q on the inner surface of the sphere and +q on its outer surface.
2. Key Action (Removing -q and Grounding): The induced charge −q on the inner surface is removed. Simultaneously, the sphere is connected to the ground.
3. Consequence of Grounding: Grounding the sphere fixes its potential at zero.
4. Source of the Potential: The potential at the sphere's center (which is equal to the potential of the grounded conductor itself) arises from two sources:
   • The point charge +q at a distance a from the center.
   • The induced charge q′ that now resides on the outer surface of the sphere. Since the sphere is a conductor, this charge q′ will be uniformly distributed, and it acts as if it were a point charge at the center for points outside the sphere (like the center point O).
5. Superposition Principle: The total potential at the center O is the sum of the potentials due to these two charges. For this potential to be zero, the two contributions must cancel each other out.

Solving for $q'$:

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

 

[TSny]

Maybe just stuck at part a with induced charge. For part b I referred to this in my teachers powerpoint.

This problem is not the same as the one given in your first post. It appears to be totally unrelated.

You mention your teacher's powerpoint. So, I'm assuming the picture above is a PowerPoint slide. If you have a question about this slide, you should start a separate thread and ask specific questions about it there. Otherwise, it will get confusing trying to decide which of your remarks refer to the original problem and which remarks refer to this new problem.

 

[TSny]

(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.

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

Is this taken directly from your teacher's PowerPoint slides? Does it represent your teacher's solution to the problem of post #1? Are you trying to understand your teacher's solution?

Reasoning:

1. Conductor as an Equipotential: The hollow spherical shell B is a conductor. In electrostatic equilibrium, every point on and inside a conductor's material is at the same potential.
2. Potential of Shell B: This potential is constant throughout the shell's material.
3. Point O is Inside the Conductor: Point O, the center of the sphere, lies inside the conducting material of shell B (since R<R1, and the shell occupies the space from R1to R2).
4. Conclusion: Therefore, point O must be at the same potential as the shell itself.

$$V_O=V_B$$

The original problem does not concern spherical shells. So, I'm confused here.

Field from the real charge +q on sphere A:
By Gauss's law, a charge +q uniformly distributed on a shell of radius R produces a field at a point r>R as if it were a point charge at the center. Since O is at r=0, this formula doesn't apply directly. However, by symmetry, a uniform spherical shell produces zero electric field everywhere inside the shell, including at the center.

For the original problem, the charge +q represents a point charge located outside the spherical conductor, It is not a charge distributed on the surface of the sphere. The two point charges +q and -q will cause the appearance of induced charge on the suface of the sphere. This induced charge will not be uniformly distributed.