A particle of charge q=1.0×10-7Cis at the center of a spherical cavity of radius 3.0cmin a chunk of metal. Find the electric field

(a)1.5cmfrom the cavity center and

(b) anyplace in the metal.

Short Answer

Expert verified

(a) The electric field at 1.5 cm from the cavity center is4.0×106N/C.

(b) The electric field at anyplace in the metal is zero.

Step by step solution

01

Listing the given quantities

  • The charge of the particle,q=1.0×10-7C.
  • The spherical cavity of radius, r = 3.0 cm.
02

Understanding the concept of Electric field

To explain the electrostatic force between the two charges, we assume that the charges create an electric field around them. The magnitude of electric field E set up by the electric charge q at a distance r is given as,

E=q4πε0r2

Using the concept of the electric field to determine the magnitude of the electric field

03

(a) Calculations for the electric field at 1.5 cm from the cavity center

The directionof the electric field atp1is away fromq1, and its magnitude is

E=q4πε0r2=(9×109N.m2/c2)(1.0×10-7C)(0.015m)2=4.0×106N/C

The electric field at 1.5 cm from the cavity center is4.0×106N/C.

04

(b) Calculations for the electric field at any place inside the metal

From Gauss law, the charge enclosed inside the metal is zero. The pointP2is inside the metal. Therefore, electric fieldE, due to the electric charge inside the metal, is zero.

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