Explanation

(a) The situation is summarised in figure.
$BC = AD = 3cm,\ \ \ \ AB = DC = 4cm,$
So, $AC = 5cm.$
[image:2_image95.emf]
Now, potential at $A$
$V_{A} = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{B}}{AB} + \frac{1}{4\pi\varepsilon_{0}}.\frac{q_{c}}{AC} + \frac{1}{4\pi\varepsilon_{0}}.\frac{q_{D}}{AD}$
$= \frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{10 \times 10^{- 12}}{4 \times 10^{- 2}} - \frac{20 \times 10^{- 12}}{5 \times 10^{- 2}} + \frac{10 \times 10^{- 12}}{3 \times 10^{- 2}} \right\rbrack$
$= 9 \times 10^{9} \times 10^{- 10}\left\lbrack \frac{10}{4} - \frac{20}{5} + \frac{10}{3} \right\rbrack$
$= \frac{9 \times 10^{- 1} \times 11}{6}$
$= 16.5 \times 10^{- 1} = 1.65V$