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Chapter 4: Q56P (page 780)

Two oppositely charged, identical insulating spheres, each 50.0 cm in diameter and carrying a uniformly distributed charge of magnitude 250渭C, are placed 1.00 m apart center to center (Fig. P23.56). (a) If a voltmeter is connected between the nearest points (a and b) on their surfaces, what will it read? (b) Which point, a or b, is at the higher potential? How can you know this without any calculations?

Short Answer

Expert verified

a) Potential difference is 12脳106V

b) Point (a) has higher potential.

Step by step solution

01

Step 1:

Given data:

R=D/2=0.25m

Q=250渭颁

Distance between center to (br) is .075

So, the potentialVadue to both charges at (a)

localid="1664273791009" Va=kQR-kqr=kQ1R-1r=9脳109脳250脳10-6脳12.5-10.75=6.0脳106V

Now the potential Vbis equal to the negative potential is

Va=-kQ1R-1r=-6.0脳106V

So, the total potential difference by the voltmeter is

Vab=Va-Vb=6.0脳106-(-6.0脳106)=12.0脳106V

Therefore, the potential difference is 12.0脳106V

02

Step 2:

As the potential (a) is higher than the (b) as the charge is positive.

So electric field is in the direction of displacement, so the integration will produce a positive voltage.

Therefore, the point (a) has greater potential.

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