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, $$\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{+9} ~\mathrm{Nm}^{2} \mathrm{C}^{-2}$$ )
Solution
<p>The electric potential (V) at a distance (r) from a point charge (Q) is given by the formula:</p>
<p>$ V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} $</p>
<p>In this case, we know (V) and (Q), and we're asked to solve for (r). We can rearrange the formula to solve for (r):</p>
<p>$ r = \frac{1}{4\pi\epsilon_0} \frac{Q}{V} $</p>
<p>Substituting the given values into this equation gives:</p>
<p>$ 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} $</p>
<p>So, the distance of the point P from the point charge is 0.9 meters. This corresponds to 90 cm.</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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