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NEET 2005
Exam
NEET
Year
2005
Subject
Physics
Chapter
Electrostatics

NEET 2005 · Physics · Electrostatics

Two charge q_1 and q_2 are placed 30 ~cm apart, as shown in the...

Two charge q1q_1 and q2q_2 are placed 30 cm30 \mathrm{~cm} apart, as shown in the figure. A third charge q3q_3 is moved along the are of a circle of radius 40 cm40 \mathrm{~cm} from C\mathrm{C} to D\mathrm{D}. The change in the potential energy of the system is (\frac{q_3 k}{4 \pi \in_0}), where kk is: Two charge q_1 and q_2 are placed 30 ~cm apart, as shown in the figure. A third charge q_3 is moved along the are of a

Options

  1. A8q18 q_1
  2. B6q16 q_1
  3. C8q28 q_2
  4. D6q26 q_2

Correct answer

C. 8q28 q_2

Step-by-step solution

Vi=14πϵ0[q1q3(0.4)+q1q2(0.3)+q2q3(0.5)]vf=14πϵ0[q1q3(0.4)+q1q2(0.3)+q2q3(0.1)]\begin{aligned}& V_i=\frac{1}{4 \pi \epsilon_0} \left[\frac{q_1 q_3}{(0.4)}+\frac{q_1 q_2}{(0.3)}+\frac{q_2 q_3}{(0.5)}\right]\\ & v_f=\frac{1}{4 \pi \epsilon_0}\left[\frac{q_1 q_3}{(0.4)}+\frac{q_1 q_2}{(0.3)}+\frac{q_2 q_3}{(0.1)}\right] \end{aligned}Two charge q_1 and q_2 are placed 30 ~cm apart, as shown in the figure. A third charge q_3 is moved along the are of a\\⇒ΔV=Vf−V1=14πϵ0q2q3(10.1−10.5)=q2q3πϵ0(10−2)=q34πϵ0(8q2)⇒k=8q2\begin{aligned} & \Rightarrow \Delta V=V f-V_1 \\ & =\frac{1}{4 \pi \epsilon_0}q_2 q_3 \left(\frac{1}{0.1}-\frac{1}{0.5}\right)\\ & =\frac{q_2 q_3}{\pi \epsilon_0}\left(10^{-2}\right)=\frac{q_3}{4 \pi \epsilon_0}\left(8 q_2\right) \\ & \Rightarrow k=8 q_2 \end{aligned}
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