86. When S is open for a long time, the charge on C is q
i
=
E
2
C. When S is closed for a long time, the
current i in R
1
and R
2
is i = (
E
2
− E
1
)/(R
1
+ R
2
) = (3.0 V
− 1.0 V)/(0.20 Ω+ 0.40 Ω) = 3.33 A. The
voltage difference V across the capacitor is then V =
E
2
− iR
2
= 3.0 V
− (3.33 A)(0.40 Ω) = 1.67 V. Thus
the final charge on C is q
f
= V C. So the change in the charge on the capacitor is ∆q = q
f
− q
i
=
(V
− E
2
)C = (1.67 V
− 3.0 V)(10 µF) = −13 µC.