PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 22, Problem 83P

(a)

To determine

To Show:The electric field inside a solid sphere having a charge density ρ at a distance r is ρr3ε0r^

(b)

To determine

The electric field at the given points 1 and 2 as shown in the figure.

Given:

The radius of the cavity is R2 .

Explanation:

Draw a diagram to show the position of the cavity.

  PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 22, Problem 83P

The electric field inside a cavity is equal to the sum of the electric field due to the original uncut sphere and the electric field due to the sphere of the cavity but with a uniform negative charge density.

Consider a point P at position 1.

Write the expression for the net electric field at position 1.

  E1=EPR+EPb

Here, E1 is the electric field at position 1, EPR is the electric field at P due to the uncut sphere of radius R and EPb is the electric field at P due to the cavity.

Substitute ρR3ε0 for EPR and ρb3ε0 for EPb in the above expression.

  E1=ρ3ε0(Rb)=ρ3ε0(RR2)=ρR6ε0

The direction will be radially outward.

Take the centrepoint O at position 2.The electric field will be only due to the cavity as the electric field due to the uncut sphere at the centre will be zero.

Write the expression for the electric field at position 2.

  E2=EPR+EPb

Here, E2 is the electric field at position 2.

Substitute 0 for EPR and ρb3ε0 for EPb in the above expression.

  E2=ρb3ε0=ρR6ε0

Negative sign indicates that the electric field will be towards the centre of the cavity.

Conclusion:

Thus, the electric field at point 1 is ρR6ε0 and the electric field at point 2 is ρR6ε0

  

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Chapter 22 Solutions

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS

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