Electrostatics Question 496
Question: Infinite charges of magnitude q each are lying at x =1, 2, 4, 8… meter on X-axis. The value of intensity of electric field at point x = 0 due to these charges will be [J & K CET 2004]
Options:
A) $ 12\times 10^{9}qN/C $
B) Zero
C) $ 6\times 10^{9}qN/C $
D) $ 4\times 10^{9}qN/C $
Show Answer
Answer:
Correct Answer: A
Solution:
Net field at origin $ E=\frac{q}{4\pi {\varepsilon _{0}}}[ \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{4^{2}}+….\infty ] $
$ =\frac{q}{4\pi {\varepsilon _{0}}}[ 1+\frac{1}{4}+\frac{1}{16}+…..\infty ] $
$ =\frac{q}{4\pi {\varepsilon _{0}}}[ \frac{1}{1-\frac{1}{4}} ]=12\times 10^{9}qN/C $
BETA
