(i) Change in capacitance
(ii) Change in electric field
(iii) Change in electric density
Dielectric slab of thickness 5‌mm is equivalent to an air capacitor of thickness =‌

5

10

‌mm
Effective separation between the plates with air in between is =(5+0.50)‌mm=5.5‌mm
(i) Effective new capacitance
‌=200µF×‌

5‌mm

5.5‌mm

=‌

2000

11

µF
‌‌≈182µF
(ii) Effective new electric field
‌=‌

100V

5.5×10−3m

=‌

20000

1.1

‌‌≈18182V∕m
(iii) ‌

‌ New energy stored ‌

‌ Original energy stored ‌

=‌

‌ 1 2 C′V2

1 2 CV2

=‌

C′

C

=‌

10

11

New Energy density will be (‌

10

11

)2 of the original energy density =‌

100

121

of the original energy density.
[Note: If the student writes C=‌

Aε0

d

‌Cm=‌

KAε0

d

‌E′=‌

V

d

‌U=‌

1

2

ε0E2