Parallel plate condenser having a plate separation d is charged to a potential V. It is then isolated. The intensity of the separation of the plates is then doubled. The new electric field intensity is :

2 E

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E/4

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E/2

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Solution

The correct option is B E
The electric field intensity is independent of the separation between the plates. $(E = \frac{Q}{A ϵ_{0}})$
thus the electric field will remain same as before i.e E

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Parallel plate condenser having a plate separation d is charged to a potential V. It is then isolated. The intensity of the separation of the plates is then doubled. The new electric field intensity is :

The correct option is B EThe electric field intensity is independent of the separation between the plates. (E=QAϵ0) thus the electric field will remain same as before i.e E

The correct option is B E
The electric field intensity is independent of the separation between the plates. $(E = \frac{Q}{A ϵ_{0}})$
thus the electric field will remain same as before i.e E