KCET · Physics · Electrostatics
A point charge \(A\) of \(+10 \mu \mathrm{C}\) and another point charge \(B\) of \(+20 \mu \mathrm{C}\) are kept 1 m apart in free space. The electrostatic force on \(A\) due to \(B\) is \(\mathbf{F}_1\) and the electrostatic force on \(B\) due to \(A\) is \(\mathbf{F}_2\). Then
- A \(\mathbf{F}_1=-2 \mathbf{F}_2\)
- B \(\mathbf{F}_1=-\mathbf{F}_2\)
- C \(2 \mathrm{~F}_1=-\mathrm{F}_2\)
- D \(\mathbf{F}_1=\mathbf{F}_2\)
Answer & Solution
Correct Answer
(B) \(\mathbf{F}_1=-\mathbf{F}_2\)
Step-by-step Solution
Detailed explanation
Electrostatic force on each charges placed at certain distance is equal in magnitude but opposite in direction.
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