question_answer

Two tiny spheres carrying charges \[1.8\,\mu C\]and \[2.8\mu C\]are located at 40 cm apart. The potential at the mid-point of the line joining the two charges is:

A) \[3.8\times {{10}^{4}}\,V\]                  

B) \[2.1\times {{10}^{5}}V\]

C) \[4.3\times {{10}^{4}}V\]                    

D) \[3.6\times {{10}^{5}}V\]

Correct Answer: B

Solution :

(b) \[2.1\times {{10}^{5}}V\] Here, \[{{q}_{1}}=1.8\mu C=1.8\times {{10}^{-6}}C\] \[{{q}_{2}}=2.8\mu C=2.8\,\times {{10}^{-6}}C\] Distance between the two spheres \[=\text{ }40\text{ }cm\]\[=0.4\text{ }m\] For the mid-point \[{{r}_{1}}={{r}_{2}}=\frac{0.40}{2}=0.2\,m\] Potential at O, \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{q}_{1}}}{{{r}_{1}}}+\frac{{{q}_{2}}}{{{r}_{2}}} \right]\] \[=\frac{9\times {{10}^{9}}\left( 1.8+2.8 \right)\times {{10}^{-6}}}{0.2}=2.1\times {{10}^{5}}V\]