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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2026 · 23 January · Evening Shift · Physics · Electrostatics

Two shorts dipoles (A, B), A having charges ± 2 μ C and length 1 cm...

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.
Two shorts dipoles (A, B), A having charges ± 2 μ C and length 1 cm and B having charges ± 4 μ C and length 1 cm are

Options

  1. A9162×104\frac{9}{16} \sqrt{2} \times 10^{4}
  2. B92×1049 \sqrt{2} \times 10^{4}
  3. C4.52×1044.5 \sqrt{2} \times 10^{4}
  4. D9162×105\frac{9}{16} \sqrt{2} \times 10^{5}

Correct answer

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

Step-by-step solution

Two shorts dipoles (A, B), A having charges ± 2 μ C and length 1 cm and B having charges ± 4 μ C and length 1 cm are
E⃗2=−KP2r3\vec{E}_2 = -\dfrac{KP_2}{r^3} ; E⃗1=−2KP1r3\vec{E}_1 = -\dfrac{2KP_1}{r^3}
P1=2×10−6×10−2=2×10−8P_1 = 2 \times 10^{-6} \times 10^{-2} = 2 \times 10^{-8}
P2=4×10−6×10−2=4×10−8P_2 = 4 \times 10^{-6} \times 10^{-2} = 4 \times 10^{-8}
E⃗net=2×9×109×2×10−8(0.4)3i^−9×109×4×10−8(0.4)3j^\vec{E}_{net} = \dfrac{2 \times 9 \times 10^9 \times 2 \times 10^{-8}}{(0.4)^3} \hat{i} - \dfrac{9 \times 10^9 \times 4 \times 10^{-8}}{(0.4)^3} \hat{j}
E⃗net=9×109×4×10−8(0.4)3[i^−j^]\vec{E}_{net} = \dfrac{9 \times 10^9 \times 4 \times 10^{-8}}{(0.4)^3} [\hat{i} - \hat{j}]
∣E⃗net∣=9×10416(2)|\vec{E}_{net}| = \dfrac{9 \times 10^4}{16} (\sqrt{2})
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