# A Parallel Plate Capacitor is to Be Designed with a Voltage Rating 1 kv, Using a Material of Dielectric Constant 3 and Dielectric Strength About 107 Vm^−1. - Physics

Numerical

A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

#### Solution

Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V

Dielectric constant of a material, in_"r" = 3

Dielectric strength = 107 V/m

For safety, the field intensity never exceeds 10% of the dielectric strength.

Hence, electric field intensity, E = 10% of 10= 106 V/m

Capacitance of the parallel plate capacitor, C = 50 pF = 50 × 10−12 F

Distance between the plates is given by,

"d" = "V"/"E"

= 1000/10^6 = 10^-3  "m"

Capacitance is given by relation

C = (in_0in_"r""A")/"d"

Where,

A = Area of each plate

in_0 = Permittivity of free space = 8.85 xx 10^-12  "N"^-1  "C"^2  "m"^-2

∴ "A" = ("Cd")/(in_0in_"r")

= (50 xx 10^-12 xx 10^-3)/(8.85 xx 10^-12 xx 3) ≈ 19  "cm"^2

Hence, the area of each plate is about 19 cm2.

Concept: The Parallel Plate Capacitor
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Chapter 2: Electrostatic Potential and Capacitance - Exercise [Page 91]

#### APPEARS IN

NCERT Physics Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.33 | Page 91
NCERT Physics Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 33 | Page 91
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