TS EAMCET · Physics · Electrostatics
Two charges are \(+10 \mu C\) and \(-10 \mu C\) are separated by \(10 \mathrm{~cm}\). The magnitude of force acting on another charge \(5 \mu \mathrm{C}\) placed at the midpoint of the line joining the two charges will be [Use \(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\) in SI unit]
- A \(360 \mathrm{~N}\)
- B \(0 \mathrm{~N}\)
- C \(320 \mathrm{~N}\)
- D \(380 \mathrm{~N}\)
Answer & Solution
Correct Answer
(A) \(360 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{F}=\mathrm{F}_1+\mathrm{F}_2 \\ & \Rightarrow \mathrm{F}=\frac{9 \times 10^9 \times 10 \times 5 \times 10^{-12}}{\left(5 \times 10^{-2}\right)^2} \times 2 \\ & \Rightarrow \mathrm{F}=\frac{9 \times 10 \times 5 \times 10^{-3}}{25 \times 10^{-4}} \times 2…
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