There are two identical, positively charged conducting spheres fixed in space. T

Question

There are two identical, positively charged conducting spheres fixed in space. The spheres are 31.8 cm apart (center to center) and repel each other with an electrostatic force of F 0.0720 N. Then, a thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed the spheres still repel but with a force of F2 0.100 N. Using this information, find the initial charge on each sphere, q and g2 if initially q2. The Coulomb Force constant is k -1/(4TEo) -8.99x10° N miC2 Number Number

Explanation / Answer

Suppose initial charge on Sphere 1 = q1

Suppose initial charge on Sphere 2 = q2

r = distance between both charges = 31.8 cm = 0.318 m

Force between both charges = 0.0720 N

Now electrostatic force is given by:

F = k*q1*q2/r^2

q1*q2 = F*r^2/k

Using given values

q1*q2 = 0.0720*0.318^2/(9*10^9)

q1*q2 = 8.09*10^-13 C

[See that q1*q2 will be Positive as force is repulsive which means both charge has same sign].

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 = 31.8 cm, 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 (0.100*0.318^2/(9*10^9))

Q = 1.06*10^-6 C

So,

(q1 + q2)/2 = 1.06*10^-6 C

q1 + q2 = 2.12*10^-6 C

We know that

q1*q2 = 8.09*10^-13 C

q1*(2.12*10^-6 - q1) = 8.09*10^-13

q1^2 - 2.12*10^-6*q1 + 8.09*10^-13 = 0

Solving above equation (Since q1 < q2)

q1 = 4.99*10^-7 C = 0.5*10^-6 C

q2 = 1.62*10^-6 C

Please Upvote. Let me know if you have any doubt.

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