Question 293: Electrostatics & Capacitance

Question

Electric potential at a point '$$\mathrm{P}$$' due to a point charge of $$5 	imes 10^{-9} \mathrm{C}$$ is $$50 \mathrm{~V}$$. The distance of '$$\mathrm{P}$$' from the point charge is:
(Assume, $$\frac{1}{4 \pi \varepsilon_{0}}=9 	imes 10^{+9} ~\mathrm{Nm}^{2} \mathrm{C}^{-2}$$ )

Accepted Answer

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 	imes 10^9 \, 	ext{Nm}^2 	ext{C}^{-2}}{1} \frac{5 	imes 10^{-9} \, 	ext{C}}{50 \, 	ext{V}} = 0.9 \, 	ext{m} $
So, the distance of the point P from the point charge is 0.9 meters. This corresponds to 90 cm.

Answer

0.9 cm

Answer

90 cm

Answer

3 cm

Answer

9 cm

Electrostatics & Capacitance

Electric potential at a point '$\mathrm{P}' due to a point charge of 5 \times 10^{-9} \mathrm{C} is 50 \mathrm{~V}. The distance of '\mathrm{P}' from the point charge is: (Assume, \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{+9} ~\mathrm{Nm}^{2} \mathrm{C}^{-2}$ )

Choose the correct answer: