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NEET 2005 (AIPMT) · Electrostatic Potential Energy

Electrostatic Potential Energy question from NEET 2005 (AIPMT)

Q60NEET 2005 · AIPMTMedium

Electrostatic Potential Energy

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 the potential energy of the system is q34πε0k\frac{q_3}{4\pi\varepsilon_0}k where k is

Figure

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Answer: (c)

A, q1B, q2DC40305010
  1. q3q_3 stays 4040 cm from q1q_1 (it moves on a circle centred at A), so only the q2q3q_2q_3 term changes.
  2. At C: BC=302+402=50BC = \sqrt{30^2 + 40^2} = 50 cm. At D: BD=40−30=10BD = 40 - 30 = 10 cm.
  3. ΔU=q2q34πε0(10.1−10.5)=q34πε0(8q2)\Delta U = \frac{q_2q_3}{4\pi\varepsilon_0}\left(\frac{1}{0.1} - \frac{1}{0.5}\right) = \frac{q_3}{4\pi\varepsilon_0}(8q_2)
  4. So k=8q2k = 8q_2, answer (c).
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