Answer:
a)14.17V
b)32.8 x
J
c)96.9x
J
d) -64x
J
Step-by-step explanation:
Given:
Area 'A'=7.10cm² =>7.1 x
m²
voltage '
'=4.8 volt
= 2.20mm => 2.2 x
m
= 6.50mm => 6.5 x
m
a) Capacitance
before push is given by:
= εA/
=>

= 2.85 x
F
=

=> 2.85 x
x 4.8
=1.37 x
C
Capacitance
after push is given by:
= εA/
=>

= 9.66 x
F
=

=

Therefore, the potential difference between the plates
= 1.37 x
/ 9.66 x
=>14.17V
b)

32.8 x
J
c)

= 96.9x
J
d) the work required to separate the plates is given by:
workdone=
-
=> 32.8 x
J- 96.9x
J
W≈ -64x
J