UPSEE Solved Model Paper 6

Parallel Plate Capacitors are the type of capacitors which are formed by arrangement of electrodes and insulating material (dielectric)
The two conducting plates act as electrodes which are separated by a dielectric between them.
The two plates of parallel plate capacitor are of equal dimensions and is connected to power supply. The plate which is connected to positive terminal of battery acquires positive charge while the plate which is connected to negative terminal of battery acquires negative charge. Due to the attraction charges store between the plates.

The amount of electric charge stored in any of the plate of parallel plate capacitor is directly proportional to the potential difference between the two plates of Parallel Plate Capacitor. This relation can be seen as:
Q ∝ V
OR
Q = (constant) V
Q = CV
Where,
C = Capacitance of capacitor
Q = Amount of charge stored in one capacitor
V = Potential difference between the two plates
The capacitance of parallel plate capacitor depends upon
The distance d between two plates
The area A of medium between the plates
According to the gauss law, the electric field can be written as:
E =

Q

ε0A

⇒ Ed = V =

Qd

ε0d

Since we know that the capacitance is defined as V =

Q

C

, so capacitance can be written as :
C =

ε0A

d

When the plates are places very close and the area of plates are large we get the maximum capacitance
Energy stored in the capacitor
U = E =

1

2

Q2

C

We know that Q = CV
U =

1

2

(CV)2

C

=

1

2

CV2
Now CV = Q
U =

1

2

QV
Thus, U =

1

2

Q2

C

=

1

2

CV2 =

1

2

QV
Where Q = Coloumb, V - volt energy U is in joule
⇒ When plates are moved apart , then distance between the plates will increase.
So, energy decreases, charge decreases, voltage decreases and capacity also decreases.