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JEE MAIN 2021
25-02-2021 S2
Question
Two identical conducting spheres with negligible volume have 2.1 nC and -0.1 nC charges, respectively. They are brought into contact and then separated by a distance of 0.5 m . The electrostatic force acting between the spheres is $\_\_\_\_$ $\times 10^{-9} \mathrm{~N}$. $$ \left[\text { Given : } 4 \pi \varepsilon_0=\frac{1}{9 \times 10^9} \text { SI unit }\right] $$
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Solution
$\begin{aligned} & q=\frac{(2.1-0.1)}{2} n C=1 n C \\ & f=\frac{9 \times 10^9 \times 10^{-18}}{(0.5)^2}=36 \times 10^{-9}\end{aligned}$
Question Tags
JEE Main
Physics
Easy
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