Two identical metallic spheres, having unequal opposite charges are placed at a distance of 0.90 m apart. After bringing them in contact with each other, they again placed at the same distance apart. Now the force of repulsion between them is 0.025 N. Calculate the final charge on each of them.
Asked by DEVESH KUMAR | 10th Nov, 2013, 12:38: AM
Expert Answer:
Let the initial charges be q1 and q2 respectively.
After they come in contact, the charges are rearranged such that they acquire same charge.
let us say that charge on each of them is Q.
They are again brought apart at a distance of 0.9 m. Hence, the force between them will be given as
F = kQ2 / r2
0.025 = (9x109 x Q2) / 0.92
Q2 = 0.025 x 0.92 / 9x109
Q = 1.5 x 10-6 C
Let the initial charges be q1 and q2 respectively.
After they come in contact, the charges are rearranged such that they acquire same charge.
let us say that charge on each of them is Q.
They are again brought apart at a distance of 0.9 m. Hence, the force between them will be given as
F = kQ2 / r2
0.025 = (9x109 x Q2) / 0.92
Q2 = 0.025 x 0.92 / 9x109
Q = 1.5 x 10-6 C
Answered by Romal Bhansali | 10th Nov, 2013, 10:48: AM
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