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Physics Question on Thermodynamics

The value of electric potential at a distance of 9 cm from the point charge 4×10−7 C4 \times 10^{-7} \, \text{C} is
Given 14πϵ0=9×109 N m2C−2\text{Given} \, \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2}

A

4×102 V4 \times 10^2 \, \text{V}

B

44.2 V44.2 \, \text{V}

C

4.4×105 V4.4 \times 10^5 \, \text{V}

D

4×105 V4 \times 10^5 \, \text{V}

Answer

4×105 V4 \times 10^5 \, \text{V}

Explanation

Solution

Electric potential (V) due to a point charge (q) at a distance (r) is given by:

V=14πϵ0qrV = \frac{1}{4\pi\epsilon_0} \frac{q}{r}

Given: 14πϵ0=9×109 N m2 C−2\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N m}^2 \, \text{C}^{-2} q = 4 × 10-7 C r = 9 cm = 0.09 m

V=(9×109 N m2 C−2)(4×10−7 C)0.09 m=4×104 VV = (9 \times 10^9 \, \text{N m}^2 \, \text{C}^{-2}) \frac{(4 \times 10^{-7} \, \text{C})}{0.09 \, \text{m}} = 4 \times 10^4 \, \text{V}