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Chapter 26: Q. 9 (page 736)

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 Q1, is at potential V1, and the electric field strength at its surface is E1. The values for sphere 2 are Q2,V2,andE2.

a. Is V1larger than, smaller than, or equal to V2? Explain.

b. Is Q1larger than, smaller than, or equal to Q2? Explain.

c. Is E1 larger than, smaller than, or equal to E2? Explain.

Short Answer

Expert verified

a.V1=V2

b.Q1>Q2

c.E1<E2

Step by step solution

01

: Given information and formula used   

Given :

After the switch is closed,

Charge, potential and electric field strength of sphere 1 : Q1,V1,E1

Charge, potential and electric field strength of sphere 1 : Q2,V2,E2.

Theory used :

Two proportional laws are :

V鈭�Qr

E鈭�Vr

02

Determining if V1 larger than, smaller than, or equal to V2 

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

V1=V2.

03

Determining if Q1 larger than, smaller than, or equal to Q2?

(b) Because V鈭�Qr, the smaller sphere must carry less charge to have the same potential, hence Q1>Q2.

04

Determining if E1 larger than, smaller than, or equal to E2 

c. For the electric field on the sphere's surface, we know that E鈭�VrBecause both spheres have the same potential, the sphere with the smaller radius Vrwill have the greater ratio. i.e.

E1<E2.

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