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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Physics
Chapter
Electrostatics

BITSAT 2014 · Physics · Electrostatics

Two point dipoles p k and p 2 k are located at (0,0,0) and (1 m,...

Two point dipoles pk^\mathrm{p} \hat{\mathrm{k}} and p2k^\frac{\mathrm{p}}{2} \hat{\mathrm{k}} are located at (0,0,0)and(1m,0,2m)and (1 \mathrm{ m}, 0,2 \mathrm{ m}) respectively. The resultant electric field due to the two dipoles at the point (1m,0,0)(1 \mathrm{ m}, 0,0) isTwo point dipoles p k and p 2 k are located at (0,0,0) and (1 m, 0,2 m) respectively. The resultant electric field due

Options

  1. A9p32πϵ0k^\frac{9 \mathrm{p}}{32 \pi \epsilon_{0}} \hat{\mathrm{k}}
  2. B−7p32πϵ0k^\frac{-7 \mathrm{p}}{32 \pi \epsilon_{0}} \hat{\mathrm{k}}
  3. C7p32πϵ0k^\frac{7 \mathrm{p}}{32 \pi \epsilon_{0}} \hat{\mathrm{k}}
  4. DNone of these

Correct answer

B. −7p32πϵ0k^\frac{-7 \mathrm{p}}{32 \pi \epsilon_{0}} \hat{\mathrm{k}}

Step-by-step solution

Two point dipoles p k and p 2 k are located at (0,0,0) and (1 m, 0,2 m) respectively. The resultant electric field due
The given point is at axis of P→2\frac{\overrightarrow{\mathrm{P}}}{2} dipole and at equatorial line of P→\overrightarrow{\mathrm{P}} dipole so that field at given point.
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