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 placed at the same distance apart.Now the force of repulsion between them is 0.0025N.Calculate the final charge on each of them
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 placed at the same distance apart.Now the force of repulsion between them is 0.0025N.Calculate the final charge on each of them
Let the initial charges be q and q' respectively. Now, after the contact is made the charges are rearranged (as spheres are of unequal and opposite charges) such that they acquire same charge, let us say Q. Now, it is such that Q = (q - q')/2
so, after they are again brought apart at a distance of 0.9 m, the force between them will be given as
F = kQ2 / r2
or
0.025 = (9x109 x Q2) / 0.92
or
Q2 = 0.025 x 0.92 / 9x109
thus,
Q = (2.25 x 10-12)1/2
so, final charge on each of the spheres will be
Q = 1.5 x 10-6 C
Let the initial charges be q and q' respectively. Now, after the contact is made the charges are rearranged (as spheres are of unequal and opposite charges) such that they acquire same charge, let us say Q. Now, it is such that Q = (q - q')/2
so, after they are again brought apart at a distance of 0.9 m, the force between them will be given as
F = kQ2 / r2
or
0.025 = (9x109 x Q2) / 0.92
or
Q2 = 0.025 x 0.92 / 9x109
thus,
Q = (2.25 x 10-12)1/2
so, final charge on each of the spheres will be
Q = 1.5 x 10-6 C
