- #1
Terry Bing
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- 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 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.
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