Question

Two identical metallic spheres, having unequal, opposite charges are placed at a distance 0.90m apart in air. After bringing them in contact with each other,they are 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.

Answer 1

Let the initial charges q1 and q2 respectively

after charges coming in contact , they rearrange such a that they have equal or same charge.---------------(given)

let assume charge on both charges is Q

charges are again brought apart at the distance of 0.9 m.-------(given)

Hence,
the force between them is given as :

F = KQ^2/ r^2

0.025 = (9 × 10^9 × Q^2) / (0.9)^2

Q^2 =0.025 × 0.9^2 / 9×10^9 ----(calculate it by rounding off figures and basic maths)

Answer is
Q = 1.5 × 10^-6--------(one point five into ten to the bar minus six)


hope it will help you

Answer 2

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 = (9×109 x Q2) / 0.92

Q2 = 0.025 x 0.92 / 9×109

Q = 1.5 x 10-6 C