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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Physics
Chapter
Electrostatics

JIPMER 2015 · Physics · Electrostatics

Two charges of +10 μ C and +20 μ C are separated by a distance 2 cm...

Two charges of +10μC+10 \mu \mathrm{C} and +20μC+20 \mu \mathrm{C} are separated by a distance 2 cm . The net potential (electric) due to the pair at the middle point of the line joining the two changes, is

Options

  1. A27 MV
  2. B18 MV
  3. C20 MV
  4. D23 MV

Correct answer

A. 27 MV

Step-by-step solution

Using the equation V=Q4πε0IV=\frac{Q}{4 \pi \varepsilon_0 I} The potential due to +10μC+10 \mu \mathrm{C} is V1=(10×10−6)×(9×109)1×10−2=9MVV_1=\frac{\left(10 \times 10^{-6}\right) \times\left(9 \times 10^9\right)}{1 \times 10^{-2}}=9 \mathrm{MV} The potential due to +20μC+20 \mu \mathrm{C} is V2=20×10−6×9×1091×10−2=18MVV_2=\frac{20 \times 10^{-6} \times 9 \times 10^9}{1 \times 10^{-2}}=18 \mathrm{MV} The net potential at the given point is 9MV+18MV=27MV9 \mathrm{MV}+18 \mathrm{MV}=27 \mathrm{MV}
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