# Electrostatic force between a Half Cylinder and a Plate

Terry Bing

## Homework Statement

(This is not a HW problem, but HW-type problem.)
A half cylinder of radius R and length L>>R is formed by cutting a cylindrical pipe made of an insulating material along a plane containing its axis. The rectangular base of the half cylinder is closed by a dielectric plate of length L and width 2R. A charge Q on the half cylinder and a charge q on the dielectric plate are uniformly sprinkled. Find the electrostatic force between the half cylinder and the dielectric plate.

## Homework Equations

Flux through a closed surface $S$ enclosing a charge $q_{enc}$ is
$$\oint_S \vec{E} \cdot \vec{dA} \ =\frac{q_{enc}}{\epsilon _0}$$
Force $\vec{dF}$ on a charged surface carrying a surface charge density $\sigma$ due to an electric field $\vec{E}$ is
$$\vec{dF}=\sigma \vec{E} dA$$, where dA is the area element (scalar, not vector).

## The Attempt at a Solution

Brute force integration gave me
$$F=\frac{qQ}{8\epsilon_0 R L}$$
Since this problem is from a high school textbook, I think brute force double integration $\int_S \vec{dF}$ is not the expected solution. So I tried the following:
For the flat plate which has uniform charge distribution,
$$\int_S \vec{dF}=\int_S\sigma \vec{E} dA$$
By symmetry of the problem, we know that resultant force will be normal to the plate. Hence we pick the normal component of $\vec{E}$. Let the magnitude of the net force be $F$
$$\lvert \int_S \vec{dF} \rvert=\sigma \int_SE_\perp dA$$
$$\implies F=\sigma \int_S \vec{E} \cdot \vec{dA}$$
$$\implies F=\sigma \, \Phi$$
where $\Phi$ is the flux due to the half cylinder through the plate.
$$\implies F=\frac{q}{2RL} \, \Phi$$

Now, total flux emerging from the half cylinder is
$$\Phi_{tot} =\frac{Q}{\epsilon _0}$$
Out of this if $\Phi=\frac{1}{4}\Phi_{tot} =\frac{Q}{4 \epsilon _0}$ passes through the plate, then we are done. But I don't see how I can show this.
Any help would be appreciated.

Last edited:

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I don't see how I can show this.
What angle does the plate subtend at a point on the half pipe?

• TSny
Terry Bing
What angle does the plate subtend at a point on the half pipe?
Depends on the point we choose. For eg at a point adjacent to the plate it subtends zero angle.
The farthest point does subtend pi/2. If that was true for all points, then great. But that,s not true for all points, is it?

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Depends on the point we choose. For eg at a point adjacent to the plate it subtends zero angle.
No. The plate occupies a diameter of the half pipe. What angle does a diameter of a circle subtend at its circumference?

• Terry Bing
Terry Bing
No. The plate occupies a diameter of the half pipe. What angle does a diameter of a circle subtend at its circumference?
Awesome! Thanks.
I was considering the extreme case of a point on the edge of the half pipe, on which the plate seems to subtend zero angle. I guess as a limit, it would still subtend pi/2.

• TSny
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Awesome! Thanks.
I was considering the extreme case of a point on the edge of the half pipe, on which the plate seems to subtend zero angle. I guess as a limit, it would still subtend pi/2.
Right.

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What a nice problem!

@Terry Bing, This is from a high-school text? Interesting! Do you mind giving the title and author of the text?

Terry Bing
What a nice problem!

@Terry Bing, This is from a high-school text? Interesting! Do you mind giving the title and author of the text?
Sorry, not a text, a problem book. 'Advanced Problems in school physics', by cengage publication. I think it is out of publication. Problems compiled by indian authors , I think mostly physics olympiad problems.

Homework Helper
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Sorry, not a text, a problem book. 'Advanced Problems in school physics', by cengage publication. I think it is out of publication. Problems compiled by indian authors , I think mostly physics olympiad problems.
OK. Thank you.

quite simple unitary method, 4pi angle enclosed means Q/Eo. Therefore, pi angle enclosed here which means Q/4Eo

bhavesh_k_tutor

## Homework Statement

(This is not a HW problem, but HW-type problem.)
A half cylinder of radius R and length L>>R is formed by cutting a cylindrical pipe made of an insulating material along a plane containing its axis. The rectangular base of the half cylinder is closed by a dielectric plate of length L and width 2R. A charge Q on the half cylinder and a charge q on the dielectric plate are uniformly sprinkled. Find the electrostatic force between the half cylinder and the dielectric plate.

## Homework Equations

Flux through a closed surface S enclosing a charge qenc is
∮SE→⋅dA→ =qencϵ0
Force dF→ on a charged surface carrying a surface charge density σ due to an electric field E→ is
dF→=σE→dA, where dA is the area element (scalar, not vector).

## The Attempt at a Solution

Brute force integration gave me
F=qQ8ϵ0RL
Since this problem is from a high school textbook, I think brute force double integration ∫SdF→ is not the expected solution. So I tried the following:
For the flat plate which has uniform charge distribution,
∫SdF→=∫SσE→dA
By symmetry of the problem, we know that resultant force will be normal to the plate. Hence we pick the normal component of E→. Let the magnitude of the net force be F
|∫SdF→|=σ∫SE⊥dA
⟹F=σ∫SE→⋅dA→
⟹F=σΦ
where Φ is the flux due to the half cylinder through the plate.
⟹F=q2RLΦ

Now, total flux emerging from the half cylinder is
Φtot=Qϵ0
Out of this if Φ=14Φtot=Q4ϵ0 passes through the plate, then we are done. But I don't see how I can show this.
Any help would be appreciated.