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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Physics
Chapter
Electrostatics

BITSAT 2021 · Physics · Electrostatics

A charge Q is placed at each of the opposite corners of a square. A...

A charge QQ is placed at each of the opposite corners of a square. A charge qq is placed at each of the other two corners. If the net electrical force on QQ is zero, then Qq\frac{Q}{q} equals:A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two

Options

  1. A−1-1
  2. B11
  3. C−12-\frac{1}{\sqrt{2}}
  4. D−22-2 \sqrt{2}

Correct answer

D. −22-2 \sqrt{2}

Step-by-step solution

A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two
It is given that F⃗A+F⃗B+F⃗D=0\vec{F}_{A}+\vec{F}_{B}+\vec{F}_{D}=0 Where F⃗A,F⃗B\vec{F}_{A}, \vec{F}_{B} and F⃗D\vec{F}_{D} are the forces applied by charges placed at A,BA, B and DD on the charge placed at CC. \begin{array}{l} \Rightarrow \overrightarrow{\mathrm{F}_{\mathrm{B}}+\overrightarrow{\mathrm{F}_{\mathrm{D}}=-\overrightarrow{\mathrm{F}_{\mathrm{A}} \\ \left|\overrightarrow{\mathrm{F}_{\mathrm{B}}+\overrightarrow{\mathrm{F}_{\mathrm{D}}\right|=\sqrt{2} \frac{\mathrm{KqQ}}{\mathrma^{2}} \\ \Rightarrow\left|\overrightarrow{\mathrm{F}_{\mathrm{A}}\right|=\frac{\mathrm{KQ}^{2}}{2 a^{2}} \\ \therefore \sqrt{2} \frac{\mathrm{KqQ}}{a^{2}}=-\frac{\mathrm{KQ}^{2}}{2 a^{2}} \\ \Rightarrow \frac{\mathrm{Q}}{q}=-2 \sqrt{2} \end{array}
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