Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.6825 N when separated by 50 cm, center-to-center. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.1986 N. What were the initial charges on the spheres? Since one is negative and you cannot tell which is positive or negative, there are two solutions. Take the absolute value of the charges and enter the smaller value AND larger value.

Answer 1

Ans:-

Given data Fa = -0.6825N , r = 50cm = 0.5m,Fr = 0.1986N

Let the charges on the spheres be q1 and q2. If the force of attraction between them has magnitude 0.6825 N, then Coulomb’s law gives us

F = k |q1q2|/r^2

= 9*10^9*|q1q2|/0.5^2 = 0.6825

|q1q2| = 0.6825*0.5^2/9*10^9 = 1.896*10^-11C^2 = 1.9*10^-11

But since we are told that the charges attract one another, we know that q1 and q2 have opposite signs and so their product must be negative. So we can drop the absolute value sign

q1q2 = -1.9*10^-11

Then the two spheres are joined by a wire. The charge is now free to re–distribute itself between the two spheres. If the new charge on each sphere is Q, then

Q + Q = 2Q = q1 + q2

The force of repulsion between the spheres is now 0.1986 N, so that

F = KQ^2 /r^2 = 9*10^9 *Q^2/0.5^2 = 0.1986

Q^2 = 0.1986*0.5^2/9*10^9 =5.5*10^-12

We don’t know what the sign of Q is, so we can only say:

Q = (+-)2.34*10^-6

q1 + q2 = 2Q = (+-)4.8*10^-6

First, choosing the + sign

q2 = 4.8*10^-6 - q1…………………………….1

q1 *(4.8*10^-6 –q1) = -1.9*10^-11

which we can rewrite as

q1^2 – 4.8*10^-6q1 - 1.9*10^-11 = 0

which is a quadratic equation for q1. When we find the solutions; we get:

q1 = -2.58*10^-6 C 0r 7.4*10^-6 C

Putting these possibilities into 1 we find

q2= 7.38*10^-6 or -2.6*10^-6C

but these really give the same answer: One charge is −2.6*10^−6 C and the other is +7.38 * 10^−6 C.