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Physics · Electrostatics

JEE Advanced 2023 — Paper 2 — Question 14

An electric dipole is formed by two charges +q and -q located in xy-plane at (0,2)mm(0,2) \mathrm{mm} and (0,−2)mm(0,-2) \mathrm{mm}, respectively, as shown in the figure. The electric potential at point P(100,100)mmP(100,100) \mathrm{mm} due to the dipole is V0\mathrm{V}_{0}. The charges +q and -q are then moved to the points (−1,2)mm(-1,2) \mathrm{mm} and (1,−2)mm(1,-2) \mathrm{mm}, respectively. What is the value of electric potential at PP due to the new dipole?

Question figure
  1. Option A:

    V0/4V_{0} / 4

  2. Option B:

    V0/2\mathrm{V}_{0} / 2

    Correct
  3. Option C:

    V0/2\mathrm{V}_{0} / \sqrt{2}

  4. Option D:

    3V0/43 V_{0} / 4

Answer: B

Step-by-step solution

Case (i) V0=kP0r2cos⁡45∘\mathrm{V}_{0}=\frac{\mathrm{kP}_{0}}{\mathrm{r}^{2}} \cos 45^{\circ}

figure

Case (ii) Component of P0′\mathrm{P}_{0}^{\prime} along x -axis =−P02i^=-\frac{\mathrm{P}_{0}}{2} \hat{\mathrm{i}} Component of P0′P_{0}^{\prime} along yy-axis =P0j^=P_{0} \hat{j} Potential due to P0j^\mathrm{P}_{0} \hat{\mathrm{j}} ⇒kP0cos⁡45∘r2=V0\Rightarrow \frac{\mathrm{kP}_{0} \cos 45^{\circ}}{\mathrm{r}^{2}}=\mathrm{V}_{0} Potential due to −P02i^-\frac{\mathrm{P}_{0}}{2} \hat{\mathrm{i}} ⇒k(P02)cos⁡135∘r2=−V02\Rightarrow \frac{\mathrm{k}\left(\frac{\mathrm{P}_{0}}{2}\right) \cos 135^{\circ}}{\mathrm{r}^{2}}=-\frac{\mathrm{V}_{0}}{2} So, net potential due to new dipole (P0′)=V0−V02=V02\left(\mathrm{P}_{0}^{\prime}\right)=\mathrm{V}_{0}-\frac{\mathrm{V}_{0}}{2}=\frac{\mathrm{V}_{0}}{2}

figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics
Topic
Electric Dipole