Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
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# A Parallel-plate Capacitor with the Plate Area 100 Cm2 and the Separation Between the Plates 1⋅0 Cm is Connected Across a Battery of Emf 24 Volts. Find the Force of Attraction Between the Plates. - Physics

ConceptCapacitors and Capacitance

#### Question

A parallel-plate capacitor with the plate area 100 cm2 and the separation between the plates 1⋅0 cm is connected across a battery of emf 24 volts. Find the force of attraction between the plates.

#### Solution

Area of the plates of the capacitor, A = 100 cm2 = 10^-2  "m"^2

Separation between the plates, d = 1 cm = 10^-2  "m"

Emf of battery, V = 24 V

Therefore,

Capacitance , C = (∈_0A)/d = ((8.85 xx 10^-12) xx (10^-2))/(10^-2) = 8.85 xx 10^-12 V

Energy stored in the capacitor,

E = 1/2 CV^2 = 1/2 xx (8.85 xx 10^-12) xx (24)^2

= 2548.8 xx 10^-12  "J"

Force of attraction between the plates, F = E/d = (2548.8 xx 10^-12)/(10^-2) = 2548.8 xx 10^-10  "N"

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Solution A Parallel-plate Capacitor with the Plate Area 100 Cm2 and the Separation Between the Plates 1⋅0 Cm is Connected Across a Battery of Emf 24 Volts. Find the Force of Attraction Between the Plates. Concept: Capacitors and Capacitance.
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