11.3 Solutions
521
11.78 (a) Battery remains connected
(i) V = V ; potential remains unchanged.
(ii) E = E; electric field is unchanged.
(iii) q = K q; charge is increased by a factor K . The additional charge
(K − 1)q is moved from the negative to the positive plate by the
battery, as the dielectric slab is inserted.
(iv) C = K C; capacitance is increased by a factor K .
(v) U =
1
2
q V =
1
2
KqV = K U
Energy is increased by a factor K .
(b) The battery is disconnected
(i) V =
V
K
; potential is decreased by a factor K
(ii) E =
E
K
; electric field is decreased by a factor K . Both (i) and (ii)
follow from the fact that q = q so that C V = C V and V =
C V
C =
V
K
. Same reasoning holds good for E .
(iii) q = q; charge remains unchanged as there is no path for charge
transfer.
(iv) C = K C; capacitance is increased by a factor K .
(v) U =
1
2
q V =
1
2
qV
K
=
U
K
The energy is lowered by a factor K .
11.79 (a) The battery remains connected
(i) V = V ; potential remains unchanged.
(ii) E < E; the electric field is decreased since E = V /d, and V is
constant.
(iii) C < C; capacitance is reduced since C ∝ 1/d.
(iv) q < q; the charge is reduced since q = C V , with C decreasing and V remaining constant. Some charge is transferred from the
capacitor to the charging battery.
(v) U < U ; the energy is decreased since U =
1
2
qV , with q decreasing and V remaining constant.
(b) Battery is disconnected
(i) V > V , the potential increases because q = C V , with C decreasing and q remaining constant.
(ii) E = E, the electric field is constant because q = C V =
ε 0 AV
d
=
ε 0 AE, with q remaining constant.
(iii) C < C; the capacitance is decreased since C ∝ 1/d.