Question 701: Electrostatics & Capacitance

Question

The value of electric potential at a distance of $$9 \mathrm{~cm}$$ from the point charge $$4 	imes 10^{-7} \mathrm{C}$$ is [Given $$\frac{1}{4 \pi \varepsilon_0}=9 	imes 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}$$] :

Accepted Answer

The electric potential (V) at a distance (r) from a point charge (q) is given by:
$$V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$$
where $$\frac{1}{4 \pi \varepsilon_0}$$ is Coulomb's constant.
In this case, we have:
$$q = 4 	imes 10^{-7} \mathrm{C}$$
$$r = 9 \mathrm{~cm} = 0.09 \mathrm{~m}$$
$$\frac{1}{4 \pi \varepsilon_0}=9 	imes 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}$$
Substituting these values into the equation for electric potential, we get:
$$V = (9 	imes 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}) \frac{(4 	imes 10^{-7} \mathrm{C})}{(0.09 \mathrm{~m})}$$
$$V = 4 	imes 10^4 \mathrm{~V}$$
Therefore, the value of electric potential at a distance of $$9 \mathrm{~cm}$$ from the point charge $$4 	imes 10^{-7} \mathrm{C}$$ is
$$4 	imes 10^4 \mathrm{~V}$$
.
The correct answer is Option D.

Answer

$$4 	imes 10^2 \mathrm{~V}$$

Answer

$$44.4 \mathrm{~V}$$

Answer

$$4.4 	imes 10^5 \mathrm{~V}$$

Answer

$$4 	imes 10^4 \mathrm{~V}$$

Electrostatics & Capacitance

The value of electric potential at a distance of $9 \mathrm{~cm} from the point charge 4 \times 10^{-7} \mathrm{C} is [Given \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}$] :

Choose the correct answer: