Question

Two spherical conductors A and B of radius 1m and carrying a charge +4 microcoulomb and +2 microcoulomb respectively, they are placed with a distance between them a) which sphere has larger potential? b) when they are connected using a wire in which direction does the charge flows? c) what is the new charge on each spheres?

Answer 1

ANSWER:

a) SPHERE A.

b) SPHERE A TO SPHERE B.

c)$3 \times {10}^{ - 6}C$

GIVEN:

$Q_1 = 4 \times {10}^{ - 6}C \\ Q_2 = 2 \times {10}^{ - 6}C \\ r = 1 \: m$

TO FIND:

a) LARGER POTENTIAL

b) DIRECTION OF FLOW OF CHARGE WHEN CONNECTED BY A WIRE

c) NEW CHARGE AFTER CONNECTION.

FORMULAE:

$V = \frac{1}{4\pi\epsilon_o} \times \frac{Q}{r}$

EXPLANATION:

a) As V $\alpha$ Q and r is constant. So Sphere A which has higher charge will have the higher potential.

b) Charge always flows from higher potential to lower potantial. Hence it moves from Sphere A to Sphere B.

c) TO ATTAIN EQUILIBRIUM, ELECTRIC FIELD AND POTENTIAL SHOULD BE EQUAL . AS 4π, $\epsilon_o$ AND r are constant charge of Sphere A should be equal to Charge of Sphere B

$\frac{4 \times {10}^{ - 6} + 2 \times {10}^{ - 6} }{2} =( 6 \times {10}^{ - 6} ) \div 2 \\ = 3 \times {10}^{ - 6}C$

Answer 2

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