Physics, 10.03.2020 18:09 maxi12312345

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A piece of material having a dielectric constant of 3.55 is placed between the plates, completely filling the space between them. Part A. How much energy is stored in the capacitor before the dielectric is inserted?
Part B. How much energy is stored in the capacitor after the dielectric is inserted?
Part C. By how much did the energy change during the insertion?

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Answer from: tracyaleblanc

a) $3.27\cdot 10^{-3} J$

b) $11.60\cdot 10^{-3} J$

c) $8.33\cdot 10^{-3} J$

Explanation:

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

$C=13.5 \mu F = 13.5\cdot 10^{-6}F$ is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

$U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J$

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

C'=kC

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

$U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2$

where:

$C=13.5\cdot 10^{-6}F$ is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

$U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J$

The initial energy stored in the capacitor, before the dielectric is inserted, is

$U=3.27\cdot 10^{-3} J$

The final energy stored in the capacitor, after the dielectric is inserted, is

$U'=11.60\cdot 10^{-3} J$

Therefore, the change in energy of the capacitor during the insertion is:

$\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J$

So, the energy of the capacitor has increased by $8.33\cdot 10^{-3} J$

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

Answer from: Quest

The magnets are probably damaged or there not strong enough to attached to the refrigerator

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