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Two charges $q_1$ and $q_2$ are placed 30 cm apart, as shown in the figure. A third charge $q_3$ 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 $\dfrac{q_3}{4\pi\varepsilon_0}k$, where $k$ is -
($q_1$ is at A and $q_2$ at B, with AB = 30 cm. The arc is centred at A; C is 40 cm from A perpendicular to AB, and D lies on the line AB, 40 cm from A, i.e. 10 cm beyond B.)
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($q_1$ is at A and $q_2$ at B, with AB = 30 cm. The arc is centred at A; C is 40 cm from A perpendicular to AB, and D lies on the line AB, 40 cm from A, i.e. 10 cm beyond B.)
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A
$8q_2$
B
$6q_2$
C
$8q_1$
D
$6q_1$
Detailed Solution
Initially $q_3$ is at C: AC = 0.4 m, and BC $= \sqrt{0.4^2 + 0.3^2} = 0.5$ m.
$U_i = \dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{q_1 q_3}{0.4} + \dfrac{q_1 q_2}{0.3} + \dfrac{q_2 q_3}{0.5}\right]$
Finally $q_3$ is at D: AD = 0.4 m and BD = 0.4 $-$ 0.3 = 0.1 m.
$U_f = \dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{q_1 q_3}{0.4} + \dfrac{q_1 q_2}{0.3} + \dfrac{q_2 q_3}{0.1}\right]$
The distance of $q_3$ from $q_1$ does not change (it moves on a circle centred at A), so only the $q_2q_3$ term changes.
$\Delta U = U_f - U_i = \dfrac{q_2 q_3}{4\pi\varepsilon_0}\left[\dfrac{1}{0.1} - \dfrac{1}{0.5}\right] = \dfrac{q_2 q_3}{4\pi\varepsilon_0}(10 - 2)$
$\Delta U = \dfrac{q_3}{4\pi\varepsilon_0}(8q_2)$
$\Rightarrow k = 8q_2$
$U_i = \dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{q_1 q_3}{0.4} + \dfrac{q_1 q_2}{0.3} + \dfrac{q_2 q_3}{0.5}\right]$
Finally $q_3$ is at D: AD = 0.4 m and BD = 0.4 $-$ 0.3 = 0.1 m.
$U_f = \dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{q_1 q_3}{0.4} + \dfrac{q_1 q_2}{0.3} + \dfrac{q_2 q_3}{0.1}\right]$
The distance of $q_3$ from $q_1$ does not change (it moves on a circle centred at A), so only the $q_2q_3$ term changes.
$\Delta U = U_f - U_i = \dfrac{q_2 q_3}{4\pi\varepsilon_0}\left[\dfrac{1}{0.1} - \dfrac{1}{0.5}\right] = \dfrac{q_2 q_3}{4\pi\varepsilon_0}(10 - 2)$
$\Delta U = \dfrac{q_3}{4\pi\varepsilon_0}(8q_2)$
$\Rightarrow k = 8q_2$