AP EAMCET · PHYSICS · Electrostatics
Two charges \(Q\) and \(4 Q\) are separated by a distance of \(6 \mathrm{~cm}\). The distance of the point from \(4 Q\) at which net electric field is zero is
- A \(2 \mathrm{~cm}\)
- B \(6 \mathrm{~cm}\)
- C \(8 \mathrm{~cm}\)
- D \(4 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(A) \(2 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Let the electric field be zero at a point \(P\). \(\begin{aligned} & \frac{K Q}{x^2}=\frac{K 4 Q}{(6-x)^2} \quad\left(\because E=\frac{K Q}{r^2}\right) \\ & (6-x)^2=4 x^2 \Rightarrow 6-x=2 x \\ & \Rightarrow x=2 \mathrm{~cm}\end{aligned}\)
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