Question

A 200 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 thickness 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 of the capacitor is $C = 181.8 \mu F$

Electric field between the plates is $E = 10^4 N/C$

Energy density of the field is $u = 4.42 \times 10^{-4} J/m^3$

Explanation:

As we know that capacitance of parallel plate capacitor is given as

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

so we have

$200\mu F = \frac{\epsilon_o A}{d}$

now when the distance between the plates is double and a dielectric slab of half thickness is inserted between the plates

so we have

$C' = \frac{\epsilon_o A}{\frac{t_1}{k_1} + \frac{t_2}{k_2}}$

now we will have

$C' = \frac{\epsilon_o A}{\frac{d}{10} + d}$

$C' = \frac{200\mu F}{1.1} = 181.8 \mu F$

Now we know that electric field is defined as

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

$E = \frac{100}{0.010}$

$E = 10^4 N/C$

Now we know that energy density is given as

$u = \frac{1}[2}\epsilon_o E^2$

$u = \frac{1}[2}(8.85 \times 10^{-12})(10^4)^2$

$u = 4.42 \times 10^{-4} J/m^3$

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Topic : Capacitor

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Answer 2

Answer:

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