Question

Two identical metallic balls, having similar charges of
different magnitudes, when placed at a distance of 0.5 m
repel each other with a force of 0.108 N. They have been
touched with each other and again placed at the same
distance, they repel each other with a force of 0.144 N
Find the initial charge on each ball.​

Answer 1

Two identical small conducting spheres, having charges of opposite sign, attract each other with a force of 0.108N when separated by 0.5m. The spheres are connected by a conducting wire, which is then removed, and thereafter, they repel each other with a force of 0.036N. The initial charges of the spheres are.

Given:

Two identical metallic balls, having similar charges of different magnitudes when placed at a distance of 0.5 m repel each other with a force of 0.108 N. They have been touched by each other and again placed at the same distance, they repel each other with a force of 0.036 N.

To Find:

The initial charge on each ball.

Solution:

To find the initial charge on each ball we will follow the following steps:

Let the initial charges be q1 and q2.

As we know,

Electrical Force =

$\frac{kq1q2}{ {r}^{2} }$

The Force unit is the newton.

In the first case when charges are q1 and q2, r = 0.5m and force = 0.108N.

So,

$0.108 \: = \frac{9 \times {10}^{9} \times q1 \times q2}{ {(0.5)}^{2} }$

$q1 × q2 = 3 \times {10}^{ - 12}$

Now,

In the second case when charges are touched charges are divided equally because both the balls are the same and the new charge will be

$\frac{q1 + q2}{2}$

Now,

Force in the second case is:

$0.036 = \frac{k {(q1 + q2)}^{2} }{ {2}^{2} \times {(0.5)}^{2} } = 9 \times {10}^{9} \times {(q1 + q2)}^{2}$

${(q1 + q2)}^{2} = \frac{0.036}{9 \times {10}^{9} } = 4\times {10}^{ - 12}$

Also,

$q1+q2 = 2 \times {10}^{ - 6}$

So, by hit and trial method,

We can say that q1 is $= 3 \times {10}^{ - 6}C and q2 is 3 \times {10}^{ - 6}$

Henceforth, the initial charge on each ball is 3×10-6 C and 1×10-6 C.

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Answer 2

Answer: The initial charges on each balls are

$\bold{q_1=\pm 1\times 10^{-6} \ C}\\\bold{q_2=\pm 3\times 10^{-6} \ C}$

Given:

• Two identical metallic balls.
• Distance between balls is 0.5 m.
• First force of repulsion =0.108 N.
• Second force of repulsion =0.108 N.

To Find: Initial charge on each ball

Explanation:

Step 1: We shall do the following actions to determine each ball's starting charge:

Let q1 and q2 represent the initial charges.

The fact is, $F=k\frac{q_1q_2}{r^2}$N

Now put the value of r and F we get,

$q_1q_2=3\times 10^{-12}$

Step 2: Now, when charges are contacted in the second scenario, they are distributed evenly since both balls are the same and the new charge will be the same as given $=\frac{q_1+q_2}{2}$

In second case the force will be

$0.036=k\frac{(q_1+q_2)^2}{2^2\times 0.5^2} \\0.036=9\times 10^9\times (q_1+q_2)^2\\(q_1+q_2)^2=4\times 10^{-12}\\(q_1+q_2)=2\times 10^{-6} ...eq(1)$

and $q_1q_2=3\times 10^{-12} ...eq(2)$

using above two equations we can get

$q_1=\pm 1\times 10^{-6} \ C\\q_2=\pm 3\times 10^{-6} \ C$

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