Question

Two Charges of the same magnitude q1 = q2 = 10^-6 C are located at a distance r. it takes 1J of energy to move q2 away to a distance r'>>>r, connsider r' as infinity.
a) what is the sign of the charge q2 if the charge q1 is postiive?
b) what is the distance r?
c) what is Vr->r'=Vr' - Vr?
please explain the steps to get each anwser

Answer 1

Potential energy of 2 charges=-W=-1

U=k(10^-6)(10^-6)[(1/infinite)-(1/r)]=-k(10^-6)(10^-6)(1/r)

as U is negative and U=-1

so both charges should be either negative or positive as q1 is +ve

so q2 is also +ve

U=-k(10^-6)(10^-6)(1/r)=-1

r=0.9cm

Vr->r'=Vr' - Vr=0-K(q2)/r=10⁶J/C