A hollow spherical shell with a diameter of 3.8 cm has a charge of 1.91 n C placed at its center.Calculate the electric flux through a portion of the shell with an area of 1.40 X 10-2 m2zero80 Nm^2/C56 2 Nm^2/C215.8 Nm^2/C

Answered: A hollow spherical shell with a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

A hollow spherical shell with a diameter of 3.8 cm has a charge of 1.91 n C placed at its center.
Calculate the electric flux through a portion of the shell with an area of 1.40 X 10-2 m2
zero
80 Nm^2/C
56 2 Nm^2/C
215.8 Nm^2/C

Expert Solution

GIVEN:

Diameter of the hollow spherical shell is given as, $d = 3.8 c m$

Radius of hollow spherical shell is found to be, $r = \frac{d}{2}$

$r = 1.9 c m$

Charge of the shell is given as, $q = 1.91 n C$ (or) $q = 1.91 \times 10^{- 9} C$

The area of portion in the shell is given as, $d A = 1.40 \times 10^{- 2} m^{2}$

TO DETERMINE:

Electric flux through a portion of shell having area, $d A$

SOLUTION:

According to Gauss law, The electric flux for a small portion of area $d A$ is calculated using the formula,

$ϕ = \int E d A \cos θ$

Where,

$E$ is the electric field of the point charge,

$d A$ is the small area of the spherical shell

$θ$ is the angle between the electric field lines and the area

The electric field of the point charge, which is placed inside the shell is calculated using the following consideration. Let us select a Gaussian surface concentric with the shell of radius less than the radius of the hollow spherical shell. Therefore,

$\oint E . d A = E (4 {πr}^{2})$

According to Gauss law, we have,

$E (4 {πr}^{2}) = \frac{Q}{ε_{o}}$

Where,

$ε_{o}$ is the permittivity of free space.

Since, the charge is enclosed inside the spherical shell is zero, here the electric field also becomes zero.

$E = 0$

Therefore, the electric flux also becomes zero.

$ϕ = 0$

Hence, OPTION (A) is Correct.

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