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jok3333 [9.3K]
3 years ago
6

The dielectric in a capacitor serves two purposes. It increases the capacitance, compared to an otherwise identical capacitor wi

th an air gap, and it increases the maximum potential difference the capacitor can support. If the electric field in a material is sufficiently strong, the material will suddenly become able to conduct, creating a spark. The critical field strength, at which breakdown occurs, is 3.0 MV/m for air, but 60 MV/m for Teflon.
Part A
A parallel-plate capacitor consists of two square plates 16cm on a side, spaced 0.45mm apart with only air between them. What is the maximum energy that can be stored by the capacitor?
Part B
What is the maximum energy that can be stored if the plates are separated by a 0.45-mm-thick Teflon sheet?
Physics
1 answer:
VikaD [51]3 years ago
8 0

Answer: 580 x 10^-3 J

Explanation:

0.6mm is 0.6/1000 = 600*10^-6 m

The plate area is .17*.17 = 28.9*10^-3 m^2

Air:

The voltage that can be sustained by 0.60 mm of air dielectric is:

V = 3.0*10^6* 600*10^-6 = 1800 V

The capacitance is:

C = ε*A/d = 8.854*10^-12 * 28.9*10^-3/600*10^-6 = 426*10^-12 F = 426 pF

The energy stored in a capacitor is:

E = (1/2)*C*V^2 = (1/2)*426*10^-12*(1800)^2 = 691*10^-6 J

Teflon:

The voltage is:

V = 60*10^6* 600*10^-6 = 36*10^3 = 36 kV

According to the listed reference, the relative dielectric constant for teflon is 2.1, this figure multiplies the "ε" of free space.

The capacitance is:

C = ε*A/d = 2.1*8.854*10^-12 * 28.9*10^-3/600*10^-6 = 896*10^-12 F = 896 pF

It would have been easier to note that the capacitance is 2.1 times the air-dielectric case.

The maximum energy stored is:

E = (1/2)*C*V^2 = (1/2)* 896*10^-12* (36*10^3)^2 = 580*10^-3 J

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nlexa [21]

Answer:

t=564.83seconds=9.414minutes=0.1569hours

Explanation:

Here the steady speed of the river water relative to ground is

V_{wg}=0.55m/s

Speed of the boy relative to still water is

V_{BW}=1.50m/s

Here we are considering +ve speed along along +ve x-axis and -ve speed along -ve x-axis

Therefore the speed of boy relative to the ground upstream is:

V_{BGup}=V_{BW}-V_{WG}\\V_{BGup}=1.50m/s-0.550m/s\\V_{BGup}=0.95m/s

And the speed of boy relative to the ground downstream is:

V_{BGdown}=V_{BW}+V_{WG}\\V_{BGdown}=1.50m/s+0.550m/s\\V_{BGdown}=2.05m/s

The distance covered in upstream trip d₁=1.0km=1000m and the distance covered in downstream is d₂=1.0km=1000m

Since time taken by a person is to cover distance d with speed v given by:

t=d/v

The total time taken for one trip is:

t=\frac{d_{1} }{V_{BGup} } -\frac{d_{2} }{V_{BGdown} }\\t=\frac{1000m}{0.95m/s}-\frac{1000m}{2.05m/s}\\ t=564.83seconds=9.414minutes=0.1569hours\\

3 0
3 years ago
A 6.0-kg object, initially at rest in free space, "explodes" into three segments of equal mass. Two of these segments are observ
sladkih [1.3K]

Answer:

Q = 2000 J

Explanation:

As we know that the 6 kg object was at rest initially

So here since net force on the system is zero

so the momentum of the system will always remains conserved

so we can say

0 = P_1 + P_2 + P_3

now we know that

P_1 = P_2 = P

and the angle between the two objects is 60 degree

so we can say

\vec P_1 + \vec P_2 = \sqrt{P_1^2 + P_2^2 + 2P_1 P_2cos60}

\vec P_1 + \vec P_2 = \sqrt{P^2 + P^2 + 2P^2(0.5)} = \sqrt3 P

now we can say that the speed of the third mass will be

v_3 = \sqrt 3 (20) m/s

now the total kinetic energy released in this system is given as

Q = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 + \frac{1}{2}mv_3^2

Q = \frac{1}{2}(2)(20^2) + \frac{1}{2}(2)(20^2) + \frac{1}{2}(2)(20\sqrt3)^2

Q = 2000 J

3 0
4 years ago
A dog starts at the origin and runs forward at 6m/s for 1.5s and then turns around to fetch the ball by running backward at 7m/s
Vera_Pavlovna [14]

Answer:

Total time elapsed between the start and when he returns with the ball is 7.5s

Explanation:

From the question,

- The dog starts at the origin and runs forward at 6m/s for 1.5s. First, we will determine the distance covered while running forward.

From

Speed = Distance / Time

Distance = Speed × Time

Speed = 6m/s

Time = 1.5s

∴ Distance = 6m/s × 1.5s

Distance = 9m

That is, the dog covered a distance of 9m while running forward.

- The dog turns around and runs backward at 7m/s for 3s. Now, we will also determine the distance the dog covered backwards.

Distance = Speed × Time

Speed = 7m/s

Time = 3s

Distance = 7m/s × 3s

Distance = 21m

The dog's displacement from the origin is 21m - 9m = 12m

Now, to calculate how much time has elapsed between the start if the dog runs back to the origin at 4m/s, we will first determine the time the dog spent back to the origin and then add to the time spent for the first two distances.

To get back to the origin, the dog needs to cover 12m

From

Speed = Distance / Time

Time = Distance / Speed

Distance = 12m

Speed = 4m/s

∴ Time = (12m) / (4m/s)

Time = 3s

Therefore, the dog spent 3s to run back to the origin.

Hence, total time elapsed = 1.5s + 3s + 3s

Total time elapsed = 7.5s

8 0
3 years ago
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vesna_86 [32]

Answer:

Electric potential, E = 2100 volts

Explanation:

Given that,

Electric field, E = 3000 N/C

We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m

The electric potential is given by :

V=E\times d

V=3000\ N/C\times 0.7\ m

V = 2100 volts

So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.

6 0
3 years ago
Do objects with more mass have smaller change in velocity?
Elena-2011 [213]

the object con tinues to fall but with NNO acceleration! notice that there is not net force if the upward and dowward forces are equal.

8 0
3 years ago
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