# Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (12) QE, where Q is the charge on the capacitor, and E is the magnitude of the electric field - Physics

Numerical

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor, and E is the magnitude of the electric field between the plates. Explain the origin of the factor 1/2.

#### Solution

Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx.

Hence, work done by the force to do so = FΔx

As a result, the potential energy of the capacitor increases by an amount given as uAΔx.

Where,

u = Energy density

A = Area of each plate

d = Distance between the plates

V = Potential difference across the plates

The work done will be equal to the increase in the potential energy i.e.,

FΔx = uAΔx

"F" = "uA" = (1/2in_0"E"^2)"A"

Electric intensity is given by,

"E" = "V"/"d"

∴ "F" = 1/2 in_0("V"/"d")"EA"

= 1/2(in_0"A" "V"/"d")"E"

However, capacitance, C = (in_0"A")/"d"

∴ "F" = 1/2("CV")"E"

Charge on the capacitor is given by,

Q = CV

∴ F = 1/2"QE"

The physical origin of the factor, 1/2, in the force formula lies in the fact that just outside the conductor, the field is E, and inside it is zero.

Hence, it is the average value, "E"/2, of the field that contributes to the force.

Concept: The Parallel Plate Capacitor
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.28 | Page 89
NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 28 | Page 90
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