A) 17.2V
B) 23.6V
C) 13.6V
D) 27.2V
Correct Answer: D
Solution :
The electric potential (V) at a point due to a test charge (q) at a distance r is \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] Given, \[r=0.53\times {{10}^{-10}}m\], \[q=1.6\times {{10}^{-19}}C\] \[\therefore \] \[V=\frac{9\times {{10}^{9}}\times 1.6\times {{10}^{-19}}}{0.53\times {{10}^{-10}}}\] V = 27.2 VoltYou need to login to perform this action.
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