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Question
A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected? (a) The electric field between the plates of the capacitor Justify your answer by writing the necessary expressions.
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Solution
The electric field between the plates is
`E=V/D`
The distance between plates is doubled, d' = 2d
`E'=(V')/(D')=(V/K)xx1/(2d)=1/2(E/K)`
Therefore, if the distance between the plates is double, the electric field will reduce to one half.
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