A hollow cylinder has a charge q coulomb within it. If φ is the electric flux, in unit of V m, associated with the curved surface B, the flux linked with the plane surface A, in unit of V m, will be

Question

A hollow cylinder has a charge q coulomb within it. If φ is the electric flux, in unit of V m, associated with the curved surface B, the flux linked with the plane surface A, in unit of V m, will be (A) (1/2)((q/ε0)-φ) (B) (q/2 ε0) (C) (φ/3) (D) (q/ε0)-φ

Tardigrade Admin · Accepted Answer

Apply Gauss's law to calculate the charge associated with plane surface

A

.
Gauss's law states that the net electric flux through any closed surface is equal to the net charge inside the surface divided by

ε_{0}

.
i.e.,

ϕ_{total} = \frac{q}{ε _{0}}

Let electric flux linked with surfaces

A, B

and

C

are

ϕ_{A}, ϕ_{B}

and

ϕ_{C}

, respectively. That is

ϕ_{total} = ϕ_{A} + ϕ_{B} + ϕ_{C}

Since,

ϕ_{C} = ϕ_{A}

\Rightarrow 2 ϕ_{A} + ϕ_{B} = ϕ_{total} = \frac{q}{ε _{0}}

or

ϕ_{A} = \frac{1}{2} (\frac{q}{ε _{0}} - ϕ_{B})

But

ϕ_{B} = ϕ

(given)
Hence,

ϕ_{A} = \frac{1}{2} (\frac{q}{ε _{0}} - ϕ)

Q. A hollow cylinder has a charge $q$ coulomb within it. If $ϕ$ is the electric flux, in unit of $Vm$ , associated with the curved surface $B$ , the flux linked with the plane surface $A$ , in unit of $Vm$ , will be

$\frac{1}{2} (\frac{q}{ε _{0}} - ϕ)$

$\frac{q}{2 ε _{0}}$

$\frac{ϕ}{3}$

$\frac{q}{ε _{0}} - ϕ$

Q. A hollow cylinder has a charge q coulomb within it. If ϕ is the electric flux, in unit of Vm, associated with the curved surface B, the flux linked with the plane surface A, in unit of Vm, will be

21​(ε0​q​−ϕ)

2ε0​q​

3ϕ​

ε0​q​−ϕ

Q. A hollow cylinder has a charge $q$ coulomb within it. If $ϕ$ is the electric flux, in unit of $Vm$ , associated with the curved surface $B$ , the flux linked with the plane surface $A$ , in unit of $Vm$ , will be

$\frac{1}{2} (\frac{q}{ε _{0}} - ϕ)$

$\frac{q}{2 ε _{0}}$

$\frac{ϕ}{3}$

$\frac{q}{ε _{0}} - ϕ$