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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 24 January · Evening Shift · Physics · Electrostatics

A small uncharged conducting sphere is placed in contact with an...

A small uncharged conducting sphere is placed in contact with an identical sphere but having 4×10−8C4 \times 10^{-8} \mathrm{C} charge and then removed to a distance such that the force of repulsion between them is 9×10−3 N9 \times 10^{-3} \mathrm{~N}. The distance between them is (Take 14πϵo\frac{1}{4 \pi \epsilon_{\mathrm{o}}} as 9×109inSI9 \times 10^9 \mathrm{in} \mathrm{SI} units)A small uncharged conducting sphere is placed in contact with an identical sphere but having 4 × 10^ -8 C charge and

Options

  1. A3 cm
  2. B2 cm
  3. C4 cm
  4. D1 cm

Correct answer

B. 2 cm

Step-by-step solution

A small uncharged conducting sphere is placed in contact with an identical sphere but having 4 × 10^ -8 C charge and
F=k(θ2)(θ2)r29×10−3=9×109×(4×10−8)×4×10−84×r2r2=9×109×16×10−164×9×10−3=4×10−4r=2×10−2 m⇒2 cm\begin{aligned} & \mathrm{F}=\frac{\mathrm{k}\left(\frac{\theta}{2}\right)\left(\frac{\theta}{2}\right)}{\mathrm{r}^2} \\ & 9 \times 10^{-3}=\frac{9 \times 10^9 \times\left(4 \times 10^{-8}\right) \times 4 \times 10^{-8}}{4 \times \mathrm{r}^2} \\ & \mathrm{r}^2=\frac{9 \times 10^9 \times 16 \times 10^{-16}}{4 \times 9 \times 10^{-3}}=4 \times 10^{-4} \\ & \mathrm{r}=2 \times 10^{-2} \mathrm{~m} \Rightarrow 2 \mathrm{~cm}\end{aligned}
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