Answer:
(a) The charge on capacitor in first case is
C
(b) In second case is
C
(c) Potential across plate is 24 V
(d) Work required to pull the plates is
J
Explanation:
Given:
Area of plate
![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
Voltage
V
Separation between two plate
m
(a)
The charge on capacitor is given by,
![C = \frac{Q}{V}](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7BQ%7D%7BV%7D)
But capacitance of parallel plate capacitor is given by,
![C = \frac{\epsilon_{o} A }{d}](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B%5Cepsilon_%7Bo%7D%20A%20%7D%7Bd%7D)
⇒ ![Q = \frac{\epsilon _{o}A V }{d}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5Cepsilon%20_%7Bo%7DA%20V%20%7D%7Bd%7D)
Where ![\epsilon _{o} = 8.85 \times 10^{-12}](https://tex.z-dn.net/?f=%5Cepsilon%20_%7Bo%7D%20%3D%208.85%20%5Ctimes%2010%5E%7B-12%7D)
![Q = \frac{8.85\times 10^{-12} \times 4 \times 10^{-4} \times 12}{0.50 \times 10^{-3} }](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B8.85%5Ctimes%2010%5E%7B-12%7D%20%20%5Ctimes%204%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctimes%2012%7D%7B0.50%20%5Ctimes%2010%5E%7B-3%7D%20%7D)
C
(b)
The charge on the capacitor when plate separation is
m is,
Here
m
![Q = \frac{\epsilon _{o}A V }{d}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5Cepsilon%20_%7Bo%7DA%20V%20%7D%7Bd%7D)
![Q = \frac{8.85\times 10^{-12} \times 4 \times 10^{-4} \times 12}{1 \times 10^{-3} }](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B8.85%5Ctimes%2010%5E%7B-12%7D%20%20%5Ctimes%204%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctimes%2012%7D%7B1%20%5Ctimes%2010%5E%7B-3%7D%20%7D)
C
(c)
The potential difference across plate is,
![V = \frac{Q}{C}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7BQ%7D%7BC%7D)
But ![C = \frac{\epsilon_{o} A }{d}](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B%5Cepsilon_%7Bo%7D%20A%20%7D%7Bd%7D)
![C = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4} }{1 \times 10^{-3} }](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B8.85%20%5Ctimes%2010%5E%7B-12%7D%20%5Ctimes%204%20%5Ctimes%2010%5E%7B-4%7D%20%20%7D%7B1%20%5Ctimes%2010%5E%7B-3%7D%20%7D)
F
Put the value of capacitance and find potential difference,
![V = \frac{8.5 \times 10^{-11} }{3.45 \times 10^{-12} }](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B8.5%20%5Ctimes%2010%5E%7B-11%7D%20%7D%7B3.45%20%5Ctimes%2010%5E%7B-12%7D%20%7D)
V
(d)
Work required is given by,
![W= U_{f} - U_{i}](https://tex.z-dn.net/?f=W%3D%20U_%7Bf%7D%20-%20U_%7Bi%7D)
![W = \frac{Q (V_{f} -V_{i} )}{2}](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7BQ%20%28V_%7Bf%7D%20-V_%7Bi%7D%20%20%29%7D%7B2%7D)
Where
V and ![V_{i} = 12](https://tex.z-dn.net/?f=V_%7Bi%7D%20%3D%2012)
![W = \frac{8.5 \times 10^{-11} \times 12}{2}](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B8.5%20%5Ctimes%2010%5E%7B-11%7D%20%5Ctimes%20%2012%7D%7B2%7D)
J
Therefore, the charge on capacitor in first case is
C and in second case is
C and potential across plate is 24 V and work required to pull the plates is
J