JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
Electric potential at a point \(P\) due to a point charge of \(5 \times 10^{-9}\; C\) is \(50 \;V\). The distance of \(P\) from the point charge is ......... \(cm\) (Assume, \(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^{+9}\; Nm ^2 C ^{-2}\))
- A \(3\)
- B \(9\)
- C \(90\)
- D \(0.9\)
Answer & Solution
Correct Answer
(C) \(90\)
Step-by-step Solution
Detailed explanation
\(V _{ P }=\frac{K Q}{ r }\) \(50=\frac{9 \times 10^9 \times 5 \times 10^{-9}}{ r }\) \(r =\frac{45}{50}=\frac{9}{10}=0.9\,m =90\,cm\)
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