Electrostatic force between a Half Cylinder and a Plate

In summary, the problem involves a half cylinder and a dielectric plate with uniform charge distribution. By using the equations for flux and force, the electrostatic force between the two objects can be found. By considering the symmetry of the problem and the angle subtended by the plate, the total flux passing through the plate can be determined, leading to the final solution of F = qQ/(8ε0RL). The problem is from a high school textbook and was solved using a simpler method than brute force integration.
  • #1
Terry Bing
48
6

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

The Attempt at a Solution


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

Now, total flux emerging from the half cylinder is
[tex]\Phi_{tot} =\frac{Q}{\epsilon _0} [/tex]
Out of this if [itex]\Phi=\frac{1}{4}\Phi_{tot} =\frac{Q}{4 \epsilon _0} [/itex] passes through the plate, then we are done. But I don't see how I can show this.
Any help would be appreciated.
 
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  • #2
Terry Bing said:
I don't see how I can show this.
What angle does the plate subtend at a point on the half pipe?
 
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  • #3
haruspex said:
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?
 
  • #4
Terry Bing said:
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?
 
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  • #5
haruspex said:
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.
 
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  • #6
Terry Bing said:
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.
 
  • #7
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?
 
  • #8
TSny said:
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.
 
  • #9
Terry Bing said:
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.
 
  • #10
quite simple unitary method, 4pi angle enclosed means Q/Eo. Therefore, pi angle enclosed here which means Q/4Eo
 
  • #11
Terry Bing said:

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.

 

Related to Electrostatic force between a Half Cylinder and a Plate

1. What is electrostatic force?

Electrostatic force is the attractive or repulsive force between two electrically charged objects. It is caused by the interaction of their electric fields.

2. How is electrostatic force calculated?

The electrostatic force between two objects is calculated using Coulomb's Law, which states that the force is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

3. How does the shape of objects affect the electrostatic force between them?

The shape of objects can affect the electrostatic force between them because it can change the distribution of charge on their surfaces. For example, a half cylinder and a plate will have different charge distributions, resulting in a different electrostatic force compared to two objects with the same charge but different shapes.

4. What factors affect the strength of electrostatic force between a half cylinder and a plate?

The strength of electrostatic force between a half cylinder and a plate is affected by the magnitude and distance of their charges, as well as the shapes and orientations of the objects. The presence of other nearby charged objects can also influence the force.

5. How is the direction of electrostatic force determined between a half cylinder and a plate?

The direction of electrostatic force between a half cylinder and a plate is determined by the relative charges and positions of the objects. Like charges will repel each other, while opposite charges will attract. The direction of the force can also be affected by the orientation of the objects, as well as the presence of other charged objects nearby.

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