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Physics · Electrostatics

JEE Main 2026 — 23 January, Evening Shift — Question 30

Two shorts dipoles (A,B),A(A, B), A having charges ±2μC\pm 2 \mu \mathrm{C} and length 1 cm and BB having charges ±4μC\pm 4 \mu \mathrm{C} and length 1 cm are placed with their centres 80 cm apart as shown in the figure. The electric field at a point PP, equi-distant from the centres of both dipoles is ____\_\_\_\_ N/C.

Question figure
  1. Option A:

    9162×105\frac{9}{16} \sqrt{2} \times 10^{5}

  2. Option B:

    4.52×1044.5 \sqrt{2} \times 10^{4}

  3. Option C:

    92×1049 \sqrt{2} \times 10^{4}

  4. Option D:

    9162×104\frac{9}{16} \sqrt{2} \times 10^{4}

    Correct

Answer: D

Step-by-step solution

E→2=−KP2r3;E→1=−2KP1r3\overrightarrow{\mathrm{E}}_{2}=-\frac{\mathrm{KP}_{2}}{\mathrm{r}^{3}} ; \overrightarrow{\mathrm{E}}_{1}=-\frac{2 \mathrm{KP}_{1}}{\mathrm{r}^{3}} P1=2×10−6×10−2=2×10−8\mathrm{P}_{1}=2 \times 10^{-6} \times 10^{-2}=2 \times 10^{-8} P2=4×10−6×10−2=4×10−8\mathrm{P}_{2}=4 \times 10^{-6} \times 10^{-2}=4 \times 10^{-8} E→net 2×9×109×2×10−8(0.4)3i^−9×109×4×10−8(0.4)3j^\overrightarrow{\mathrm{E}}_{\text {net }} \frac{2 \times 9 \times 10^{9} \times 2 \times 10^{-8}}{(0.4)^{3}} \hat{\mathrm{i}}-\frac{9 \times 10^{9} \times 4 \times 10^{-8}}{(0.4)^{3}} \hat{\mathrm{j}} E→net =9×109×4×10−8(0.4)3[i^−j^]\overrightarrow{\mathrm{E}}_{\text {net }}=\frac{9 \times 10^{9} \times 4 \times 10^{-8}}{(0.4)^{3}}[\hat{\mathrm{i}}-\hat{\mathrm{j}}] ∣E→net ∣=9×10416(2)\left|\overrightarrow{\mathrm{E}}_{\text {net }}\right|=\frac{9 \times 10^{4}}{16}(\sqrt{2})

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
23 January, Evening Shift
Subject
Physics
Chapter
Electrostatics
Topic
Electric Dipole

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