PHYSICS COLLEGE

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V. Use ϵ0 = 8.85×10⁻¹² C²/N⋅m².
The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

Answers

Answer 1

Answer:

Answer:

U_eq = 1.99 * 10^(-10) J

Explanation:

Given:

Plate Area = 10 cm^2

d = 0.01 m

k_dielectric = 3

k_air = 1

V = 15 V

e_o = 8.85 * 10 ^-12 C^2 / N .m

Equations used:

U = 0.5 C*V^2 .... Eq 1

C = e_o * k*A /d .... Eq 2

U_i = 0.5 e_o * k_i*A_i*V^2 /d ... Eq 3

For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:

U_electric = 0.5 e_o * k_1*A*V^2 /2*d

U_air = 0.5 e_o * k_2*A*V^2 /2*d

The total Energy is:

U_eq = U_electric + U_air

U_eq = 0.5 e_o * k_1*A*V^2 /2*d + 0.5 e_o * k_2*A*V^2 /2*d

U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d

Plug the given values:

U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01

U_eq = 1.99 * 10^(-10) J

Related Questions

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Some radar systems detect the size and shape of objects such as aircraft and geological terrain. What is the frequency of such a system which can detect objects as small as 19.1 cm?
What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?
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A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. If the radius of the ride is 9.0 meters, what is the acceleration of the ride?4.0 m/s2
0.67 m/s2
0.075 m/s2
54 m/s2

Answers

4.0 m/s2

it's 9 squared divided by 6

the answer is B 4.0 m/s2

Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 2 seconds.

Answers

Answer:.

Required velocity = 6.26ms^-1

Explanation:

Given,

Distance, s = 450m

Time, t = 2 sec

Step 1. We obtain the distance covered within the given time under gravitational acceleration, g = 9.8ms^-2

S = ut + (1/2)gt^2. :; u = 0

: S = (1/2)gt^2

=(1/2) (9.8)(2^2)

= 19.6m

Step 2 :

We obtain the velocity using the formula.

V^2 = u^2 + 2gs.

Where u is initial velocity, v is final/ required velocity

Again u = 0

: V^2 = 2 (9.8)(19.6)

= 39.2

: V = 6.26ms^-1

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity for the entire trip is 26.5 mi/h. (a) what is the constant speed with which the car moved during the second distance d?

Answers

A distance of d is covered with 53 mile/hr initially.Time taken to cover this distance t1 = d/53 hourNext distance of d is covered with x mile hours.Time taken to cover this distance t2 = d/x hours.We have average speed = 26.5 mile / hour

= Total distance traveled/ total time taken

= $(2d)/((d)/(53)+(d)/(x)) = (2)/((1)/(53)+(1)/(x) ) = (106x)/(x+53)$

$26.5 = (106x)/(x+53) \n \n 79.5 x = 1404.5\n \n x = 17.67 miles/hour$

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Answers

Answer:

The speed is $v =8.17 m/s$

Explanation:

From the question we are told that

The angle of slant is $\theta = 37.0^o$

The weight of the toolbox is $W_t = 92.0N$

The mass of the toolbox is $m = (92)/(9.8) = 9.286kg$

The start point is $d = 4.25m$ from lower edge of roof

The kinetic frictional force is $F_f = 22.0N$

Generally the net work done on this tool box can be mathematically represented as

$Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction$

The workdone due to weigh is = $mgsin \theta * d$

The workdone due to friction is = $F_f \ cos\theta * d$

Substituting this into the equation for net workdone

$W_(net) = mgsin\theta * d + F_f \ cos \theta *d$

Substituting values

$W_(net) = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25$

$= 309.98 J$

According to work energy theorem

$W_(net) = \Delta Kinetic \ Energy$

$W_(net) = (1)/(2) m (v - u)^2$

From the question we are told that it started from rest so u = 0 m/s

$W_(net) = (1)/(2) * m v^2$

Making v the subject

$v = \sqrt{(2 W_(net))/(m) }$

Substituting value

$v = \sqrt{(2 * 309.98)/(9.286) }$

$v =8.17 m/s$

A block m1 rests on a surface. A second block m2 sits on top of the first block. A horizontal force F applied to the bottom block pulls both blocks at constant velocity. Here m1 = m2 = m.(a)
What is the normal force exerted by the surface on the bottom block? (Use the following as necessary: m and g as necessary.)

Answers

(a) The normal force exerted by the surface on the bottom block is N1 = 2mg.

Given that,

A block m1 rests on a surface.
A second block m2 sits on top of the first block.
A horizontal force F applied to the bottom block pulls both blocks at constant velocity. Here m1 = m2 = m.

Based on the above information, we can say that the N1 is 2mg.

Learn more: brainly.com/question/17429689

Answer:

N = 2mg

Explanation:

Assuming the surface is horizontal

The surface must provide enough normal force to prevent the masses from accelerating in the vertical direction.

A wheel 2.45 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.30 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s, find the following.a. the angular speed of the wheel and, for point P
b. the tangential speed.
c. the total acceleration.
d. the angular position.