The correct option (A) 8q2
Explanation:
change in potential energy = final PE – initial PE
AC = 40 cm
AB = 30 cm
BC = 50 cm
∴ BD = 10 cm
initial PE = k[{(q1q3)/(AC)} + {(q2q3)/(BC)} + {(q1q2)/(AB)}]
Final PE = k[{(q1q3)/(AC)} + {(q2q3)/(BD)} + {(q1q2)/(AB)}]
∴ ΔPE = k[{(q1q3)/(AC)} + {(q2q3)/(BD)} + {(q1q2)/(AB)} – {(q1q3)/(AC)} – {(q2q3)/(BC)} – {(q1q2)/(AB)}]
= k[{(q2q3)/(BD)} – {(q2q3)/(BC)}]
= kq2q3[(1/0.1) – (1/0.5)]
= kq2q3[10 – 2]
ΔPE = 8kq2q3 = [8/(4π∈0)]q2 q3
given ΔPE = [q3/(4π∈0)]k
∴ [8/(4π∈0)]q2 q3 = [q3/(4π∈0)]k
∴ 8q2 = k