Question

# 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 electric field between the plates. Explain the origin of the factor .

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Solution

## Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of △x. Hence, work done by the force to do so =F△xAs a result, the potential energy of the capacitor increases by an amount given as uA△xWhere,u= Energy densityA= Area of each plated= Distance between the platesV= Potential difference across the platesThe work done will be equal to the increase in the potential energy i.e.,F△x=uA△xF=uA=(12∈0E2)A Electric intensity is given by,E=Vd∴F=12∈0(Vd)EA=12(∈0AVd)EHowever, capacitance, C=∈0Ad∴F=12(CV)ECharge on the capacitor is given by,Q=CV∴F=12QEThe physical origin of the factor, 12, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, E2, of the field that contributes to the force.

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