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Standard XI

Physics

Potential Energy of a System of Multiple Point Charges

Two charges q...

Question

Two charges $q_{1}$ and $q_{2}$ are located $30 cm$ apart, as shown in the figure. A third charge $q_{3}$ is moved along tthe arc of a circle of radius $40 cm$ from $C$ to $D$ . The change in the potential energy of the system is $\frac{n q_{2} q_{3}}{4 π ϵ_{0}}$ . Find $n$ .

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Solution

$U_{i} = \frac{1}{4 π ϵ_{0}} [\frac{q_{1} q_{3}}{(0.4)} + \frac{q_{1} q_{2}}{(0.3)} + \frac{q_{2} q_{3}}{(0.5)}]$
$U_{f} = \frac{1}{4 π ϵ_{0}} [\frac{q_{1} q_{3}}{0.4} + \frac{q_{1} q_{2}}{0.3} + \frac{q_{2} q_{3}}{0.1}]$
$Δ U = U_{f} - U_{i} = \frac{q_{2} q_{3}}{4 π ϵ_{0}} (\frac{1}{0.1} - \frac{1}{0.5})$
$= \frac{q_{3}}{4 π ϵ_{0}} (8 q_{2})$

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Two charges $q_{1}$ and $q_{2}$ are placed 30cm apart, as shown in figure. A third charges $q_{3}$ 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 $\frac{q_{3}}{4 π ϵ_{0}} k$ , where k is

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Two charges q1 and q2 are located 30 cm apart, as shown in the figure. A third charge q3 is moved along tthe arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is nq2q34πϵ0. Find n.

Ui=14πϵ0[q1q3(0.4)+q1q2(0.3)+q2q3(0.5)] Uf=14πϵ0[q1q30.4+q1q20.3+q2q30.1] ΔU=Uf−Ui=q2q34πϵ0(10.1−10.5) =q34πϵ0(8q2)

Two charges $q_{1}$ and $q_{2}$ are located $30 cm$ apart, as shown in the figure. A third charge $q_{3}$ is moved along tthe arc of a circle of radius $40 cm$ from $C$ to $D$ . The change in the potential energy of the system is $\frac{n q_{2} q_{3}}{4 π ϵ_{0}}$ . Find $n$ .