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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2021
  • Medium
  • Numerical

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 \(.......... \, \times 10^{-9} \,N\) [Given : \(4 \pi \varepsilon_{0}=\frac{1}{9 \times 10^{9}} SI\) unit]

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Answer & Solution

Correct Answer

(C) \(36\)

Step-by-step Solution

Detailed explanation

\(q =\frac{(2.1-0.1)}{2}\, nC =1\, nC\)

\(f =\frac{9 \times 10^{9} \times 10^{-18}}{(0.5)^{2}}=36 \times 10^{-9}\)

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