Question

2 identical metal spheres having charges of opposite sign attract each other with a force of 0.108N,when seperated by 0.5m. these spheres are brought in contact with each other and then seperated . Then they repel each other with a force 0.036N. what were the initial charges?

Answer 1

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Answer 2

Hello friend...!!

According to the given question,
• F1 = 0.108.
• F2 = 0.036.
• r = 0.5m.

Let us consider the charges on the sphere q1 and q2 .

So after connection we know that the charges gets redistributed in sphere, so let q be the charge on each sphere.

q1 + q2 = 2q ----(1) { conservation of charges }

We know that,
F1 = kq1q2 / r^2 ---(2)

Also, F2 = kq^2 / r^2 .-----(3)
Now, 0.036 = 9x10^9 x q^2 / (0.5)^2.
From this we can able to find q

Therefore, q = 0.036x (0.5)^2 / 9x10^9.

Therefore , q = 1 x 10^-12
Implies q = 1 x 10^-6 C. -----(4)

F1/F2 = q1q2 / q^2 .
3 = q1q2 / q^2.
3q^2 = q1q2.
Now,
(q1 - q2)^2 = (q1+q2)^2.
(2q)^2 - 3q^2/r^2 .
Implies q^2/r^2.

q1 - q2 = q/r .
1x 10^-6 / 0.5 = 2x 10^-6.-----(5).

Also,
q1 + q2 = 4 x 10^-6.------(6)

Now, by adding equation (5)&(6)we get,
Q1 = 3 x 10^-6

And by subracting equation (5)from(6) gives,
Q2 = 1 x 10^-6.

Therefore the initial charges are q1= 3x10^-6 and q2 = 1x10^-6.

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Hope it is useful...!!