- #1

JoelKTH

- 29

- 1

- Homework Statement
- Two large metal plates form a 60-degree angle with each other. On the bisector of the angle at a distance a from the apex, there is a point charge q. Determine the force on the charge!

- Relevant Equations
- $F = qE$

\(E = \frac{q(r - r')}{4\pi\epsilon_0 |r - r'|^3}\)

Hi,

I'm having some trouble understanding how to solve this problem. I have a few questions:

1. I understand that I need to make an educated guess for the electric potential, where \(V_1\) is given by:

V_1 = \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{r_3} - \frac{1}{r_4} + \frac{1}{r_5} - \frac{1}{r_6}\right) V_2 Is this guess chosen because it satisfies \(\nabla^2 V_1 = 0\)? Is this due to the fact that \(V_2\) at the metal plate has no potential? Could you explain the reasoning behind this educated guess? Also, I am a bit confused as for when \(V_1\) becomes significant close to the real charge, why is \(r_1 \rightarrow 0\)? Can you elaborate on this point? I know that \(V_1 = V_2\) at the boundary, but I'm not entirely clear on how the educated guess applies in this context.

2. According to the figure, the electric field in the air from the real surface charge is the same as the electric field from the 5 fictive mirror charges. How can I determine the geometry of these fictive charges? In simpler cases, like planes, I understand that if there is a positive charge \(q\) followed by a metal surface, I can place a fictitious charge of \(-q\) below and calculate the electric field. However, given the complex symmetry in this problem, I'm quite lost. How can I set up this problem correctly?

3. I've attached two pictures illustrating how the solution gives me a tip for how the problem behaves, but I have some uncertainties there.

4. When I attempt to find the distance between \(q\) and the fictive charges using the method of mirroring, I get confused.

Finally, I'm aware that \(F = qE\).

I try to calculate the electric field \(E = \frac{q(r - r')}{4\pi\epsilon_0 |r - r'|^3}\) first.

Try to find the E-field and do superposition. First I need to know the distances. I have attached my thinking.

\[

\tan(30^\circ) = \frac{q}{b_1} \rightarrow b_1 = \frac{a}{\tan(60^\circ)} = \sqrt{3}a

\]

\[

\sin(60^\circ) = \frac{q}{b_2} \rightarrow b_2 = \frac{a}{\sin(60^\circ)}

\]

\[

\cos(30^\circ) = \frac{a}{b_3} \rightarrow b_3 = \frac{a}{\cos(30^\circ)} = \frac{a}{\sqrt{3}/2} = \frac{2a}{\sqrt{3}}

\]

\[

E = \frac{q(r - r')}{4\pi\epsilon_0 |r - r'|^3}

\]

Any insights or guidance you can provide would be greatly appreciated.

I'm having some trouble understanding how to solve this problem. I have a few questions:

1. I understand that I need to make an educated guess for the electric potential, where \(V_1\) is given by:

V_1 = \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{r_3} - \frac{1}{r_4} + \frac{1}{r_5} - \frac{1}{r_6}\right) V_2 Is this guess chosen because it satisfies \(\nabla^2 V_1 = 0\)? Is this due to the fact that \(V_2\) at the metal plate has no potential? Could you explain the reasoning behind this educated guess? Also, I am a bit confused as for when \(V_1\) becomes significant close to the real charge, why is \(r_1 \rightarrow 0\)? Can you elaborate on this point? I know that \(V_1 = V_2\) at the boundary, but I'm not entirely clear on how the educated guess applies in this context.

2. According to the figure, the electric field in the air from the real surface charge is the same as the electric field from the 5 fictive mirror charges. How can I determine the geometry of these fictive charges? In simpler cases, like planes, I understand that if there is a positive charge \(q\) followed by a metal surface, I can place a fictitious charge of \(-q\) below and calculate the electric field. However, given the complex symmetry in this problem, I'm quite lost. How can I set up this problem correctly?

3. I've attached two pictures illustrating how the solution gives me a tip for how the problem behaves, but I have some uncertainties there.

4. When I attempt to find the distance between \(q\) and the fictive charges using the method of mirroring, I get confused.

Finally, I'm aware that \(F = qE\).

I try to calculate the electric field \(E = \frac{q(r - r')}{4\pi\epsilon_0 |r - r'|^3}\) first.

Try to find the E-field and do superposition. First I need to know the distances. I have attached my thinking.

\[

\tan(30^\circ) = \frac{q}{b_1} \rightarrow b_1 = \frac{a}{\tan(60^\circ)} = \sqrt{3}a

\]

\[

\sin(60^\circ) = \frac{q}{b_2} \rightarrow b_2 = \frac{a}{\sin(60^\circ)}

\]

\[

\cos(30^\circ) = \frac{a}{b_3} \rightarrow b_3 = \frac{a}{\cos(30^\circ)} = \frac{a}{\sqrt{3}/2} = \frac{2a}{\sqrt{3}}

\]

\[

E = \frac{q(r - r')}{4\pi\epsilon_0 |r - r'|^3}

\]

Any insights or guidance you can provide would be greatly appreciated.

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