Question

Two conducting spheres have identical radii Initially they have charges of opposite sign and unequal magnitudes with the magnitude of the positive charge larger than the mag- nitude of the negative charge. They attract each other with a force of 0.231 N when sepa rated by 0.5 m Initial The spheres are suddenly connected by a thin conducting wire, which is then removed. Connected Now the spheres repel each other with a force of 0.022 N Final What is the magnitude of the positive charge? Answer in units of C 017 (part 2 of 2) 10.0 points What is the negative charge? Answer in units of C

Answer 1

Suppose initial charge on Sphere 1 = q1

Suppose initial charge on Sphere 2 = q2

r = distance between both charges = 0.5 m

Force between both charges = 0.231 N

Now electrostatic force is given by:

F = k*q1*q2/r^2

q1*q2 = F*r^2/k

Using given values

q1*q2 = 0.231*0.5^2/(9*10^9)

q1*q2 = -6.41667*10^-12 C

[See that q1*q2 will be negative as force is attractive which means one of them is negative and other one is positive].

Now when both spheres are brought into contact, after that charge will be equally distributed. Now charge on each sphere will be Q, where

Q = (q1 + q2)/2

Now when returned to distance r = 0.5 m, force will be repulsive because both charge will have same sign either positive or negative, So

F1 = k*Q*Q/r^2

Q^2 = F1*r^2/k

Q = sqrt (0.022*0.5^2/(9*10^9))

Q = 7.81736*10^-7 C

So,

(q1 + q2)/2 = 7.81736*10^-7 C

q1 + q2 = 1.563472*10^-6 C

We know that

q1*q2 = -6.41667*10^-12 C

q1*(1.563472*10^-6 - q1) = -6.41667*10^-12

Solving above equation

q1 = 3.43*10^-6 C (Part A ans)

q2 = -1.87*10^-6 C (Part B Ans)

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