Question

A point charge Q is located on the axis of a disc of radius R at a distance b from the plane of the disc (figure). Show that if one-fourth of the electric flux from the charge passes through the disc, then R=√3b.

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Solution

The total flux produced by the charge is, according to Gauss’s law, is Qenclosedϵ0. Only one quarter
of this flux passes through the disk. The flux through the disk is given by:

ϕdisk=∫E.dA

where the integration cover the entire area of the disk. Evaluating this integral and set it equal to Q4ϵ0 relates b to R. As shown in figure, dA is the area of annular ring with radius s and width ds. The electric field at the ring make an angle θ with the normal to to the ring, the flux through the ring is:

dϕring=E.dA=EdAcosθ=E(2πsds)cosθ

The magnitude of the electric field has the same value at all points on the ring:

Ering=Q4πϵ0r2=Q4πϵ0(s2+b2)

and, cosθ=b√s2+b2

Therefore, dϕring=Qb2ϵ0s(s2+b2)3/2ds

To get the flux through the entire disk we integrate dϕring from s = 0 to s = R

ϕdisk=∫R0dϕring

After integration, ϕdisk=Q2ϵ0[1−b√R2+b2]

Since the flux through the disk is already given by Q4ϵ0, then:

Q4ϵ0=Q2ϵ0[1−b√R2+b2]

which will give

4b2=R2+b2

Therefore, R=√3b

ϕdisk=∫E.dA

where the integration cover the entire area of the disk. Evaluating this integral and set it equal to Q4ϵ0 relates b to R. As shown in figure, dA is the area of annular ring with radius s and width ds. The electric field at the ring make an angle θ with the normal to to the ring, the flux through the ring is:

dϕring=E.dA=EdAcosθ=E(2πsds)cosθ

The magnitude of the electric field has the same value at all points on the ring:

Ering=Q4πϵ0r2=Q4πϵ0(s2+b2)

and, cosθ=b√s2+b2

Therefore, dϕring=Qb2ϵ0s(s2+b2)3/2ds

To get the flux through the entire disk we integrate dϕring from s = 0 to s = R

ϕdisk=∫R0dϕring

After integration, ϕdisk=Q2ϵ0[1−b√R2+b2]

Since the flux through the disk is already given by Q4ϵ0, then:

Q4ϵ0=Q2ϵ0[1−b√R2+b2]

which will give

4b2=R2+b2

Therefore, R=√3b

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