Aparallel-plate capacitor has a capacitance of 10 mf and charged with a 20-v power supply. the power supply is then removed and a dielectric material of dielectric constant 4.0 is used to fill the space between the plates. how much energy is now stored by the capacitor?

After placing the dielectrics between the plates of the capacitor, the energy stored in the capacitor becomes 0.5 J.

Capacitance is a term used to define the amount of energy stored in the form of an electric charge in an electric device.

Given data:

The capacitance of parallel-plate capacitor is, C = 10 mF = 0.01 F.

The potential of power supply is, V' = 20 V.

The dielectric constant of the material is, ∈ = 4.0.

Let us first calculate the charge stored in the capacitor as,

After placing the dielectric, the capacitance of the capacitor is,

Solving as,

Now, the expression for the energy stored in the capacitor is,

Solving as,

Thus, we can conclude that the energy stored in the capacitor is of 0.5 J.

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0.5 J

Explanation:

For this capacitor we have:

is the capacitance

V = 20 V is the potential difference

So the charge stored in the capacitor is

Later, the power supply is removed, so the charge on the capacitor will remain the same. A dielectric of dielectric constant

k = 4.0

is inserted in the gap between the plates. The capacitance of the capacitor change as follows:

C' = k C = (4.0)(0.01 F) = 0.04 F

The energy stored in the capacitor is given by

and using Q = 0.2 C, we find

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weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards