Question

The potential energy where ( q_{3} )
is at point ( mathrm{C} )
[
left.U_{1}=frac{1}{4 pi varepsilon_{0}} mid frac{q_{1} q_{3}}{0.40}+frac{q_{2} q_{3}}{sqrt{(0.40)^{2}+(0.30)^{2}}}right]
]
The potential energy when ( q_{3} ) is at point ( D )
[
begin{array}{l}
quad U_{2}=frac{1}{4 pi varepsilon_{0}}left[frac{q_{1} q_{3}}{0.40}+frac{q_{2} q_{3}}{0.10}right] quad^{q}[because cdots .
text { Thus change in potential }
begin{array}{ll}
text { energy is }
Delta U=U_{2}-U_{1} & q_{4}
end{array}
end{array}
]
( Rightarrow frac{q_{3}}{4 pi varepsilon_{0}} k=frac{1}{4 pi varepsilon_{0}}left[frac{q_{1} q_{3}}{0.40}+frac{q_{2} q_{3}}{0.10}-frac{q_{1} q_{3}}{0.40}-frac{q_{2} q_{3}}{0.50}right] )
( Rightarrow k=frac{5 q_{2}-q_{2}}{0.50}=frac{4 q_{2}}{0.50}=8 q_{2} )

# 17. 02 Two charges q, and 92 are placed 30 cm apart, as shown in the figure. A third charge 40cm 93 is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is a where k is- (1) 822 (2) 67,[AIPMT-2005] (3) 891 (4) 691 93 k ,

Solution