2.28 Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) 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 ½.

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alok kumar singh | Contributor-Level 10

3 months ago

2.28 Let F be the force applied to separate the plates of a parallel plate capacitor and let $Δ x$ be the distance.

Hence work done by the force = F $Δ x$

The potential energy increase in the capacitor = uA $Δ x$ , where u = Energy density, A = area of each plate.

If d = distance between the plates and V = potential difference across the plates, the

Work done = increase in the potential energy.

Therefore,

F $Δ x$ = uA $Δ x$ or F = uA = ( $\frac{1}{2} ε_{0} E^{2}$ )A

Electric intensity is given by

E = $\frac{V}{d}$

F = ( $\frac{1}{2} ε_{0} E$ )EA= ( $\frac{1}{2} ε_{0} \frac{V}{d}$ )EA

Since Capacitance, C = $\frac{ε_{0} A}{d}$

F = ( $\frac{1}{2} C V$ E) = $\frac{1}{2}$ QE

2.28 Let F be the force applied to separate the plates of a parallel plate capacitor and let <math><mi>Δ</mi><mi>x</mi><mi></mi></math> be the distance.Hence work done by the force = F <math><mi></mi><mi>Δ</mi><mi>x</mi></math>The potential energy increase in the capacitor = uA <math><mi></mi><mi>Δ</mi><mi>x</mi></math> , where u = Energy density, A = area of each plate.If d = distance between the plates and V = potential difference across the plates, theWork done = increase in the potential energy.Therefore,F <math><mi></mi><mi>Δ</mi><mi>x</mi></math> = uA <math><mi></mi><mi>Δ</mi><mi>x</mi></math> or F = uA = ( <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><msup><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> )AElectric intensity is given byE = <math><mfrac><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mi>d</mi></mrow></mrow></mfrac></math>F = ( <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>E</mi></math> )EA= ( <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mfrac><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mi>d</mi></mrow></mrow></mfrac></math> )EASince Capacitance, C = <math><mfrac><mrow><mrow><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>A</mi></mrow></mrow><mrow><mrow><mi>d</mi></mrow></mrow></mfrac></math>F = ( <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mi>C</mi><mi>V</mi></math> E) = <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></math> QE

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2.28 Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) 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 ½.

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