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Q. 9

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Found in: Page 736

### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651

# The two metal spheres in Figure Q26.9 are connected by a metal wire with a switch in the middle. Initially the switch is open. Sphere 1, with the larger radius, is given a positive charge. Sphere 2, with the smaller radius, is neutral. Then the switch is closed. Afterward, sphere 1 has charge , is at potential , and the electric field strength at its surface is . The values for sphere 2 are . a. Is larger than, smaller than, or equal to ? Explain. b. Is larger than, smaller than, or equal to ? Explain. c. Is larger than, smaller than, or equal to ? Explain.

a.

b.

c.

See the step by step solution

## Step 1 : Given information and formula used

Given :

After the switch is closed,

Charge, potential and electric field strength of sphere 1 :

Charge, potential and electric field strength of sphere 1 : .

Theory used :

Two proportional laws are :

## Step 2 : Determining if  larger than, smaller than, or equal to

(a) When the switch is closed, the charge is transferred from Sphere 1 to Sphere 2 until their potentials are equal, i.e.

.

## Step 3 : Determining if  larger than, smaller than, or equal to ?

(b) Because , the smaller sphere must carry less charge to have the same potential, hence .

## Step 4 : Determining if  larger than, smaller than, or equal to

c. For the electric field on the sphere's surface, we know that Because both spheres have the same potential, the sphere with the smaller radius will have the greater ratio. i.e.

.