Two identical conducting spheres, fixed in place, attract each other with an ele

Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -1.0112 N when separated by 50 cm, center-to-center. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.1165 N. What were the initial charges on the spheres? Since one is negative and you cannot tell which is positive or negative, there are two solutions. Take the absolute value of the charges and enter the smaller value here. And the larger value here.

Explanation / Answer

From Coloumb's Law

F=KQ1Q2/r2

Since two charges attract each other

-1.0112 =(9*109)Q1Q2/0.52

Q1Q2=-2.8*10-11---------1

Then the two spheres are joined by a wire ,if the new charge on each sphere is Q

Q+Q=2Q=q1+q2

Force of repulsion

0.1165=(9*109)*Q2/0.52

Q=1.8*10-6C

Q1+Q2=3.6*10-6

=>Q2=3.6*10-6-Q1

From 1

Q1(3.6*10-6-Q1)=-2.8*10-11

Q12-(3.6*10-6)Q1-2.8*10-11=0

Q1=7.4*10-6C and -3.8*10-6C

If Q1=-3.8 uC

then Q2=-(2.8*10-11)/(-3.8*10-6)=7.4 uC

if Q1=7.4uC

then

Q2=-3.8 uC

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