A) 0.44 m
B) 0.65 m
C) 0.556 m
D) 0.350 m
E) 0.5 m
Correct Answer: C
Solution :
\[{{F}_{1}}={{F}_{2}}\] \[\Rightarrow \] \[\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{5\times {{10}^{-11}}\times q}{{{(0.2+x)}^{2}}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2.7\times {{10}^{-11}}\times q}{{{x}^{2}}}\] \[\Rightarrow \] \[x=0.556\text{ }m\] from IInd charge.You need to login to perform this action.
You will be redirected in
3 sec