How is the capacitance
changed when the slab is inserted between the plates Solution Let  E0 = V0/d be the electric field between the plates when
there is no dielectric and the potential difference is V0 If the dielectric
is now inserted, the electric field in the dielectric will be E = E0/K The potential difference will then be
0
0
1
3
(
)
(
)
4
4
E
V
E
d
d
K
=
+
0
0
1
3
3
(
)
4
4
K4
E d
V
K
+K
=
+
=
The potential difference decreases by the factor (K + 3)/4K while the
free charge Q0 on the plates remains unchanged