Question

A200 microfarad parallel plate capacitor having plate separation of 5mm is charged by a100v dc source. it remains connected to the source. using an insulted handle, the distance between the plates is doubles and a dielectric sab of 5mm and dielectric constant 10is introduced between the plates. explain with reason how the capacitance, electric field, energy density of plates will change

Answer 1

Answer:

Capacitance will decrease to

$C' = 181.8\mu F$

if the distance between the plates is doubled so the electric field will become half

if the electric field intensity is decreased to half then energy density will decrease by 1/4 times

Explanation:

As we know that capacitance of parallel plate capacitor is

$C = \frac{\epsilon_0 A}{d}$

Now if the distance between the plates is doubled and dielectric is inserted between the plates

so we have

$C' = \frac{\epsilon_o A}{d + \frac{d}{k}}$

$C' = \frac{\epsilon_o A}{d} (\frac{1}{1 + \frac{1}{k}})$

$C' = \frac{200 \mu F}{1 + \frac{1}{10}}$

$C' = 181.8\mu F$

So capacitance will decrease

Now we know that electric field between the two plates is given as

$E = \frac{\Delta V}{d}$

here we know that potential difference between the plates is constant due to applied battery

So if the distance between the plates is doubled so the electric field will become half

Similarly energy density is given as

$u = \frac{1}{2}\epsilon_0 E^2$

so here if the electric field intensity is decreased to half then energy density will decrease by 1/4 times

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Topic : Capacitance

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