Answer:
[tex]\huge\boxed{\sf V = 90,000 \ v}[/tex]
Explanation:
Given data:
Charge = Q = 3 × 10⁻⁶ C
Distance = r = 0.3 m
k = 9 × 10⁹ Nm/C²
Required:
Electric Potential = V = ?
Formula:
[tex]\displaystyle V = \frac{kQ}{r}[/tex]
Solution:
Put the given data in the above formula.
[tex]\displaystyle V=\frac{(9 \times 10^9)(3 \times 10^{-6})}{0.3} \\\\V=\frac{27 \times 10^3}{0.3} \\\\V= 90 \times 10^3\\\\V = 90,000 \ v\\\\\rule[225]{225}{2}[/tex]
