A) 4N
B) 8N
C) 12N
D) zero
Correct Answer: D
Solution :
[d] When we add \[-\text{ }4\mu C\]charge to each \[{{q}^{'}}_{1}=(4-4)\mu C\] = 0 \[{{q}^{'}}_{2}=(2-4)\mu C\] \[=-2\mu C\] As \[{{F}^{'}}\propto {{q}^{'}}_{1}{{q}^{'}}_{2}\] We have, \[{{F}^{'}}=0\]You need to login to perform this action.
You will be redirected in
3 sec