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Question

# Two charges q1 and q2 are placed 30 cm apart as shown in the figure. A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D. The change in potential energy of the system is xq34πϵ0 where x is

A
6q2
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B
6q1
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C
8q2
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D
8q1
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Solution

## The correct option is C 8q2 Given that, AB=30 cm=0.3 m AC=40 cm=0.4 m Initially, q3 is at point C. BC = √AB2+AC2 ⇒BC = √0.32+0.42 ⇒BC=0.5 m The initial potential energy of the system when charge q3 was at C, Ui = k{q1q2AB+q1q3AC+q2q3BC} ⇒Ui = 14πϵ0{q1q20.3+q1q30.4+q2q30.5} Now, final potential energy of the system when charge q3 is moved from C to D, Uf = 14πϵ0{q1q30.4+q1q20.3+q2q30.1} So, Uf − Ui = 14πε0{q2q30.1−q2q30.5} ⇒Uf − Ui = 104πε0{q2q31−q2q35} ⇒Uf − Ui = 104πϵo×4q2q35 ⇒Uf − Ui =8q2q34πϵ0=xq34πϵ0 [given] ⇒x=8q2 Hence, option (a) is correct.

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