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/
=>
![((8.85*10^-^1^2)(7.1*10^-^4))/(2.2*10^-^3)](https://img.qammunity.org/2021/formulas/physics/college/awngufuxd473170v9susfkzv06roxefz7v.png)
= 2.85 x
F
=
![C_o](https://img.qammunity.org/2021/formulas/business/college/19yw7rwvqx5raft3lt6jfk4qkblhckbqky.png)
=> 2.85 x
x 4.8
=1.37 x
C
Capacitance
after push is given by:
= εA/
=>
![((8.85*10^-^1^2)(7.1*10^-^4))/(6.5*10^-^3)](https://img.qammunity.org/2021/formulas/physics/college/pyg090ouz343ad3s68dto1eac8tjiru6i7.png)
= 9.66 x
F
=
![q_1](https://img.qammunity.org/2021/formulas/chemistry/high-school/p2880j1jj5riinx5upr23zzytnq9wz4b9s.png)
=
![C_1](https://img.qammunity.org/2021/formulas/chemistry/college/b2pjg4zryx5n73axaylgeibzq1pprakvkt.png)
Therefore, the potential difference between the plates
= 1.37 x
/ 9.66 x
=>14.17V
b)
![U_i=(1)/(2)C_oV_o^2 => (1)/(2) (2.85*10^-^1^2)(4.8^2)](https://img.qammunity.org/2021/formulas/physics/college/3vcjensni6r1balndbjwuab6u5blx5ryjo.png)
32.8 x
J
c)
![U_f=(1)/(2)C_1V_1^2 => (1)/(2) (9.66*10^-^1^3)(14.17^2)](https://img.qammunity.org/2021/formulas/physics/college/5rlxbgzkxrnnen2smw3mdery59ih3pjphy.png)
= 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