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A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d. A battery fully charges the capacitor and is then disconnected. The plate separation is now increased to a distance of 2d. What would be the change, if any, of the voltage across the capacitor, the electric field between the plates, and the energy stored in the capacitor?

Respuesta :

Answer:

The answer is "Option D".

Explanation:

Please find the complete question in the attached file.

As plate separation increased to 2d the capacitance get halred but the change remain same

[tex]\therefore V=\frac{Q}{C}[/tex]

The voltage doubles are now electric field remain same because both the distance and voltage get doubled.

[tex]\to E=\frac{v}{d}\ = \frac{2v}{2d}\\\[/tex]

So,

[tex]energy=\frac{1}{2}\ \frac{Q^2}{C}\\\\c'=\frac{C}{2}\\\\E'=2E[/tex]

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