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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Electrostatics

JIPMER 2016 · Physics · Electrostatics

The charges on two spheres are +7 μ C and -5 μ C respectively. They...

The charges on two spheres are +7μC+7 \mu \mathrm{C} and −5μC-5 \mu \mathrm{C} respectively. They experience a force FF. If each of them is given an additional charge of −2μC-2 \mu \mathrm{C}, then the new force of attraction will be

Options

  1. AF
  2. BF2\frac{F}{{2}}
  3. CF3\frac{F}{\sqrt{3}}
  4. D2F

Correct answer

A. F

Step-by-step solution

 Given, q1=+7gμC=+7×10−6Cq2=−5μC=−5×10−6C\begin{aligned} & \text{ Given, } q_1=+7 g \mu \mathrm{C}=+7 \times 10^{-6} \mathrm{C} \\ & q_2=-5 \mu \mathrm{C}=-5 \times 10^{-6} \mathrm{C}\end{aligned} New force F′=F^{\prime}= ? We know that, F=14πε0⋅q1q2r2F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}F=14πε0(+7×10−6)(−5×10−6)r2=−14πε035×10−12r2NF′=14πε0(+5×10−6)(−7×10−6)r2=−14πε035×10−12r2N⇒\begin{aligned} F & =\frac{1}{4 \pi \varepsilon_0} \frac{\left(+7 \times 10^{-6}\right)\left(-5 \times 10^{-6}\right)}{r^2} \\ & =-\frac{1}{4 \pi \varepsilon_0} \frac{35 \times 10^{-12}}{r^2} \mathrm{ N} \\ F^{\prime} & =\frac{1}{4 \pi \varepsilon_0} \frac{\left(+5 \times 10^{-6}\right)\left(-7 \times 10^{-6}\right)}{r^2} \\ & =-\frac{1}{4 \pi \varepsilon_0} \frac{35 \times 10^{-12}}{r^2} \mathrm{ N} \\ & \Rightarrow\end{aligned}F′=FF^{\prime}=F
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