JEE MAIN - Physics (2023 - 8th April Evening Shift - No. 11)
Electric potential at a point '$$\mathrm{P}$$' due to a point charge of $$5 \times 10^{-9} \mathrm{C}$$ is $$50 \mathrm{~V}$$. The distance of '$$\mathrm{P}$$' from the point charge is:
(Assume, $$\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{+9} ~\mathrm{Nm}^{2} \mathrm{C}^{-2}$$ )
Explanation
The electric potential (V) at a distance (r) from a point charge (Q) is given by the formula:
$ V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} $
In this case, we know (V) and (Q), and we're asked to solve for (r). We can rearrange the formula to solve for (r):
$ r = \frac{1}{4\pi\epsilon_0} \frac{Q}{V} $
Substituting the given values into this equation gives:
$ r = \frac{9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2}}{1} \frac{5 \times 10^{-9} \, \text{C}}{50 \, \text{V}} = 0.9 \, \text{m} $
So, the distance of the point P from the point charge is 0.9 meters. This corresponds to 90 cm.
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