(adsbygoogle = window.adsbygoogle || []).push({}); A slab of copper is thrust into parallel plate capacitor as shown in the figure.

a) What is the capacitance after the slab is introduced?

This is like two capacitors in series so

[tex] C_{1} = \frac{\epsilon_{0} A}{d-b-x} [/tex]

[tex] C_{2} = \frac{\epsilon_{0} A}{x} [/tex]

when added it yields [tex] \frac{epsilon_{0} A}{d-b} [/tex]

Find the ratio of the stored energy before and after the slab is inserted if the voltage is kept constant

dividing U1 and U2 which are

[tex] U_{1} = \frac{1}{2} \frac{\epsilon_{0} A}{d-b} \Delta V^2 [/tex]

[tex] U_{2} = \frac{1}{2} \frac{\epsilon_{0} A}{d} \Delta V^2 [/tex]

[tex] \frac{U_{1}}{U_{2}} = \frac{d}{d-b} [/tex]

Find the work done on the slab as it is inserted. Is it pulled in or pushed in?

Well find the difference in the energy U1 - U2 right?

[tex] U_{f} - U_{i} = \epsilon_{0} A (\frac{1}{d-b} - \frac{1}{d} [/tex]

this change is positive value. So the slab must be pushed in. Is this right?

Please do advise on any mistakes i may have made.

Thank you for your help!

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# Homework Help: Dielectric in a capacitor

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