Question

Two identical metal spheres attract each other with a force of 7.00 × 10−6 N when they are 3.00 cm apart. The spheres are then touched together and then removed to the original separation where now a force of repulsion of 2.00 × 10−6 N is observed. What is the charge on each sphere after touching and before touching?

Answer 1

Let charges on spheres = Q1 and -Q2

then, Electric force is given by,

F = k*Q1*Q2/R^2

given, F = 7.00*10^-6 N

R = disance between spheres = 3.00 cm = 3*10^-2 m

k = 9*10^-9

So, 7.00*10^-6 = 9*10^9*Q1*Q2/(3*10^-2)^2

Q1*Q2 = 7.00*10^-6/(1*10^13)

Q1*Q2 = 7.00*10^-19 eq(1)

Remember whenever metal spheres are touched, the charge is equally distributed between them. now

Charge on bothe metal sphere = Q1' = Q2' = (Q1-Q2)/2

So, new electrical force will be,

F' = k*Q1'*Q2'/R^2

here, F' = 2.00*10^-6 N

2.00*10^-6 = (9*10^9)*((Q1-Q2)/2)^2/(3*10^-2)^2

((Q1-Q2)/2)^2 = 2.00*10^-19

(Q1 - Q2)^2 = 8*10^-19 eq(2)

From eq(1) and (2),

Q1 = 13.96*10^-10 C & Q2 = 5.01*10^-10 C

Q1 = charge on sphere 1 before touching = 13.96*10^-10 C

Q2 = charge on sphere 2 after touching = -5.01*10^-10 C

Q1' = Q2' = charge on each sphere after touching = (13.96 - 5.01)*10^-10/2 = 4.475*10^-10 C

Let me know if you've any query.