1. Write any one use of capacitor 2. Obtain an expression for capacit

1.	1. Write any one use of capacitor 2. Obtain an expression for capacitance of a parallel plate capacitor3. The Capacity of a parallel plate capacitor becomes 10µF when air between the plates is replaced by a dielectric slab (k = 2). What is the capacity of the capacitor with air in between the plates?
Answer» 1. one use of capacitor: Capacitor is used to store electric charges It is used to prevent dc current. 2. Expression for capacitance of a capacitor: Potential difference between two plates V = Ed = \(\frac{\sigma}{\varepsilon_0}d\) (∴\(E=\frac{\sigma}{\varepsilon_0}\)) V = \(\frac{Q}{A\varepsilon_0}d\) ..........(1) \((σ=\frac{Q}{A})\) Capacitance C of the parallel plate capacitor, C = \(\frac{Q}{V}\) ...........(2) Sub eq. (1) in eq. (2) C = \(\frac{Q}{\frac{Q}{A\varepsilon_0}d}\) C = \(\frac{A\varepsilon_0}{d}\) 3. C = 10µF When dielectric slab is placed, New capacitance C¹ = KC 10 × 10^-6 = 2 × C, C = 5 × 10^-6 F.

1. Write any one use of capacitor 2. Obtain an expression for capacitance of a parallel plate capacitor3. The Capacity of a parallel plate capacitor becomes 10µF when air between the plates is replaced by a dielectric slab (k = 2). What is the capacity of the capacitor with air in between the plates?

Answer»

1. one use of capacitor:

2. Expression for capacitance of a capacitor: Potential difference between two plates V = Ed

= \(\frac{\sigma}{\varepsilon_0}d\) (∴\(E=\frac{\sigma}{\varepsilon_0}\))

V = \(\frac{Q}{A\varepsilon_0}d\) ..........(1) \((σ=\frac{Q}{A})\)

Capacitance C of the parallel plate capacitor,

C = \(\frac{Q}{V}\) ...........(2)

Sub eq. (1) in eq. (2)

C = \(\frac{Q}{\frac{Q}{A\varepsilon_0}d}\)

C = \(\frac{A\varepsilon_0}{d}\)

3. C = 10µF

When dielectric slab is placed, New capacitance

C¹ = KC

10 × 10^-6 = 2 × C, C = 5 × 10^-6 F.