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2 years ago

You insert a dielectric into an air-filled capacitor. How does this affect the energy stored in the capacitor?.

1 answer:

8 0

The effect that the energy stored in the capacitor is that option b: Energy stored in the capacitor will decrease.

<h3>How does electric field affects the energy stored in a capacitor that contains a dielectric?</h3>

If a dielectric is said to be inserted into any kind of an isolated and charged capacitor, the stored energy is known to often decreases to about 33% of its real value.

The capacitance of a give group of charged parallel plates is known to often increased by the inputting of a dielectric material. The capacitance is known to be inversely proportional to the electric field that tends to exist between the plates, and thus, the presence of the any form of dielectric tend to lower the effective electric field.

Therefore, The effect that the energy stored in the capacitor is that option b: Energy stored in the capacitor will decrease.

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You insert a dielectric into an air-filled capacitor. How does this affect the energy stored in the capacitor?

a. Energy stored in the capacitor will remain same.

b. Energy stored in the capacitor will decrease.

c. Energy stored in the capacitor will increase.

d. Energy stored in the capacitor will increase first and then it will decrease.

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An archer releases an arrow toward a target. The arrow travels 166 meters in 2 seconds. The speed of the arrow is

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Explanation:

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3 years ago

Show that y = '(t - s)f(s)ds is a solution to my" + ky = f(t). Use g' (0) = 1/m and mg" + kg = 0. 6.1) Derive y' 6.2) Using g(0)

Gre4nikov [31]

Answer:

Explanation:

Given that:

$y = \int^t_og'(t-s) f(s) ds \ \text{is solution to } \ my"ky= f(t)$

where;

$g'(0) = \dfrac{1}{m}$ and $mg"+kg = 0$

$\text{Using Leibniz Formula to prove the above equation:}$

$\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt$

So, $y = \int ^t_0 g' (t-s) f(s) \ ds$

$\text{By differentiation with respect to t;}$

$y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\ y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds$

$y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)$

$Since \ \ mg" (t) +kg (t) = 0 \\ \\ \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\ put \ t \ =0 \ we \ get;\\g" (0) = - \dfrac{k}{m } g(0) \\ \\ g"(0) = 0 \ \ \ \ ( because \ g(0) =0) \\ \\$

$Now \ differentiating \ equation (111) \ with \ respect \ to \ t \\ \\ g"'(t) = -\dfrac{k}{m}g(t) \\ \\ replacing \ it \ into \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\ y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\ my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}$

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3 years ago

A pure metal can be strengthening by addition of impurities. Identify strengthening mechanism that utilize impurities and discus

tiny-mole [99]

Explanation:

Strengthening is the method to improve the strength of material because different type of material property requires for different requirement .Some times requires more ductility,some time requires more strength.

When impurities are mixed with the pure metal then the strength of metal will increases and this is called the strengthening of metal.

The methods of strengthening are as follows

1.Solid solution strengthening and alloying

2.Work hardening

3.Transformation hardening

4.Grain boundary strengthening

5.Precipitation hardening

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4 years ago

What is the short description for Determining Deflections by Using Discontinuity Functions ?

maksim [4K]

Answer? somey

Explanation:

soy sause

6 0

3 years ago

A 150-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local grav

kozerog [31]

Answer:

a)Wt =25.68 lbf

b)Wt = 150 lbf

F= 899.59 N

Explanation:

Given that

$g = 5.48 ft/s^2.$

m= 150 lbm

Weight on the spring scale(Wt) = m g

We know that

$1\ lbf=32.17 \ lmb.ft/s^2$

Wt = 150 x 5.48/32 lbf

Wt =25.68 lbf

On the beam scale

This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.

Wt = 150 lbf

If the plane is moving upward with acceleration 6 g's then the for F

F = m a

We know that

$1\ ft/s^2= 0.304\ m/s^2$

$5.48\ ft/s^2= 1.66\ m/s^2$

a=6 g's

$a=9.99\ m/s^2$

F = 90 x 9.99 N

F= 899.59 N

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3 years ago