(iii) energy density of the capacitor get affected? Justify your answer in each case.
(i) Change in Capacitance
(ii) Change in Electric field
(iii) Change in Energy density
Dielectric slab of thickness 4‌mm=‌

4

5

‌mm
=0.8‌mm
Effective (air) plate separation
=(4+0.8)‌mm=4.8‌mm
(i) Effective new capacitance
C′=100µF×‌

4‌mm

4.8‌mm

=‌

400

4.8

µF
Change on capacitor remains same
Q=C×V=100×10−6×200 =2×10−2C
New p.d. V′=‌

Q

C′

=‌

2×10−2×4.8×10−6

400

=2.4×10−6‌ volt ‌
(ii) Effective new Electric Field,
E′‌=‌

2.4×10−10

4.8×10−3

‌=5×10−8Vm−1
(iii) ‌

‌ New energy stored ‌

‌ Original energy stored ‌

=‌

‌ 1 2 C′V2

1 2 CV2

‌

C′

C

=‌

400

4.8×100

=0.8
New energy density will be (0.8) of the original density.