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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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