Explanation

(a) Consider the charge distribution as shown. Considering the branch on upper side, we have
[image:2_image121.emf]
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{q}{V_{x} - V_{A}} = 4 \times 10^{- 6}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{q\ }{V_{A} - V_{y}} = \ 2 \times 10^{- 6}$
Here, $V_{x} = 6\ volt,\ V_{y} = 0$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \frac{q}{6 - V_{A}} = 4 \times 10^{- 6}\ $ …(i)
$\ \ \ \ \ \ \ \ \ \ \ \frac{q}{V_{A} - 0} = 2 \times 10^{- 6}$ …(ii)
From Eqs. (i) and (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \frac{V_{A}}{6 - V_{A}} = 2$
$\therefore\ \ \ \ \ \ \ \ \ \ \ V_{A} = 4volt$
Similarly for the lower side branch
$\ \ \ \ \ \ \ \frac{q''}{6 - V_{B}} = \ 2 \times 10^{- 6}$ …(iii)
$\ \ \ \frac{q''}{V_{B} - 0} = \ 4 \times 10^{- 6}$ ...(iv)
From Eqs. (iii) and (iv)
$\ \ \frac{V_{B}}{6 - V_{B}} = \frac{1}{2}$
$\therefore\ \ V_{B} = 2\ volt$
$\therefore V_{A} - V_{B} = 4 - 2 = 2\ volt$