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The two plates are now to be pushed together to a separation of d/2. The pushing together can be done either with the battery connected or with it disconnected. Which way would result in the greater electric field magnitude, and by what factor

Sagot :

Answer:

Electric field will be greater when the battery is connected by the factor of 2.

Explanation:

Solution:

I will be doing some algebraic calculations to answer this question:

As we know that,

Q = CV

and

C = [tex]\frac{AE_{0} }{d}[/tex]

So, when separation = d/2, then,

[tex]C^{'}[/tex] = [tex]\frac{AE_{0} }{d/2}[/tex] by rearranging we get

So,

[tex]C^{'}[/tex]  = 2C

We further know that, Voltage will remain same if the battery is  connected.

This further implies that,

Q = CV

So,

[tex]Q^{'}[/tex] = [tex]C^{'}[/tex]V

[tex]Q^{'}[/tex]  = 2CV

[tex]Q^{'}[/tex] = 2Q

and we also know that,

Electric field E = [tex]\frac{Q}{AE_{0} }[/tex]

So, the new E or [tex]E^{'}[/tex] = [tex]\frac{Q^{'} }{AE_{0} }[/tex]

Hence,

[tex]E^{'}[/tex] = [tex]\frac{2Q}{AE_{0} }[/tex] = 2E

[tex]E^{'}[/tex] = 2E

when battery is disconnected, Q remain the same.

So,

When disconnected

E = E

E = [tex]\frac{Q}{AE_{0} }[/tex]  = Same

Hence, we can see that the magnitude of the electric does not depend upon the distance of separation. Instead it does depend upon the magnitude of charge.

So, when battery is disconnected, Q is same, so the Electric field.

But when it is connected, [tex]Q^{'}[/tex] = 2Q and the [tex]E^{'}[/tex] = 2E

So,

[tex]\frac{E connected}{E disconnected} = \frac{E^{'} }{E}[/tex] = [tex]\frac{2E}{E}[/tex] = 2

Electric field will be greater when the battery is connected by the factor of 2.

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