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.