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Karnataka Board PUCPUC Science Class 11

Find the Flux of the Electric Field Through a Spherical Surface of Radius R Due to a Charge of 10−7 C at the Centre and Another Equal Charge at a Point 2r Away from the Centre in the Following Figure.

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Question

Find the flux of the electric field through a spherical surface of radius R due to a charge of 10−7 C at the centre and another equal charge at a point 2R away from the centre in the following figure.

Answer in Brief
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Solution

Given:-

Let charge Q be placed at the centre of the sphere and Q' be placed at a distance 2R from the centre.

Magnitude of the two charges  = 10−7 C

According to Gauss's Law, the net flux through the given sphere is only due to charge Qthat is enclosed by it and not by the charge Q' that is lying outside.

So, only the charge located inside the sphere will contribute to the flux passing through the sphere.

Thus,

`phi = ∫  vec"E" . vec("d"."s") = "Q"/∈_0 = 10^-7/( 8.85 xx 10^-12)`

`=> phi = 1.1 xx 10^4   "N""m"^2   "C"^-1`

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Chapter 30: Gauss’s Law - Exercises [Page 141]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 30 Gauss’s Law
Exercises | Q 7 | Page 141
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