## Parallel plate capacitor

- A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance shown below in the figure 3.

- Suppose two plates of the capacitor has equal and opposite charge Q on them. If A is the area of each plate then surface charge density on each plate is

σ=Q/A

- We have already calculated field between two oppositely charged plates using gauss's law which is

E=σ/ε_{0}=Q/ε_{0}A

and in this result effects near the edges of the plates have been neglected.

- Since electric field between the plates is uniform the potential difference between the plates is

V=Ed=Qd/ε_{0}A

where , d is the separation between the plates.

- Thus, capacitance of parallel plate capacitor in vacuum is

C=Q/V=ε_{0}A/d (3)

- From equation 3 we see that quantities on which capacitance of parallel plate capacitor depends i.e.,ε
_{0} , A and d are all constants for a capacitor.

- Thus we see that in this case capacitance is independent of charge on the capacitor but depends on area of it's plates and separation distance between the plates.

**Question**

A parallel plate capacitor has the capacitance of 20 μ F where the distance between the plates is 16 cm.If the distance between the plates is reduced to 4 cm,its capacitance will be

(a) 20 μ F

(b) 5 μ F

(c) 60 μ F

(d) 80 μ F

**Solution**

$C \alpha \frac {1}{d}$

Therefore

$\frac {C_1}{C_2} = \frac {16}{4}=4$

or

$C_1 = 4 C_2= 80$ μ F

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