Electrostatics Question 412
Question: The potential at a point, due to a positive charge of $ 100\mu C $ at a distance of 9m, is [KCET (Med.) 2000]
Options:
A) $ 10^{4} $ V
B) $ 10^{5} $ V
C) $ 10^{6} $ V
D) $ 10^{7} $ V
Show Answer
Answer:
Correct Answer: B
Solution:
By using $ V=9\times 10^{9}\times \frac{Q}{r} $
$ =9\times 10^{9}\times \frac{100\times {{10}^{-6}}}{9}=10^{5}V $
