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Rom4ik [11]
3 years ago
8

We want to design a cylindrical vacuum capacitor, with a given radius a for the outer cylindrical shell, that will be able to st

ore the greatest amount of electrical energy per unit length, subject to the constraint that the electric field strength at the surface of the inner cylinder may not exceed E0. What radius b should be chosen for the inner cylindrical conductor, and how much energy can be stored per unit length
Physics
1 answer:
anastassius [24]3 years ago
8 0

Solution :

a). Using Gauss's law :

  $E=\frac{Q}{4 \pi \epsilon_0r^2}$  ,    $b    .........(1)

Let $E=E_0,\ r=b$ in equation (1)

Therefore, $Q=4 \pi \epsilon_0b^2E_0$  .............(2)

$V_b-V_a = \int^a_b \vec E. d\vec l$

             $=\int^a_b E \ dx$

            $=\frac{Q}{4 \pi \epsilon_0} \int^a_b \frac{1}{x^2} \ dx$

            $=\frac{Q(a-b)}{4 \pi \epsilon_0 a b}$  ....................(3)

Therefore, $U=\frac{1}{2}Q \Delta V$

                     $=\frac{1}{2}(4 \pi \epsilon_0 b^2 E_0)\left(\frac{Q(a-b)}{4\pi \epsilon_0 a b}\right)$

                     $=\frac{4 \pi \epsilon_0}{2a} \ E^2_0 b^3(a-b)$  .............(4)

Now differentiating the equation (4) w.r.t. 'b', we get

$b=\frac{3}{4}a$  

Thus the radius for the inner cylinder conductor is $b=\frac{3}{4}a$

b). For the energy storage, substitute the radius in (4), we get

$U = 4 \pi\epsilon_0 \frac{27a^3E^2_0}{512}$

This is the amount of energy stored in the conductor.

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kari74 [83]

Answer:

(A) Constant acceleration will be a=1.479m/sec^2

(B) Velocity of the car when it passes 0.14 km is 20.349 m/sec

Explanation:

It is given that driver starts from rest so initial velocity of the driver u = 0 m/sec

Distance traveled s = 0.14 km = 140 m

Time taken to cross 0.14 km is 8 sec

So time taken t = 8 sec

(A) From second equation of motion s=ut+\frac{1}{2}at^2

So 140=0\times 8+\frac{1}{2}\times a\times 8^2

a=1.479m/sec^2

(B) From third equation of motion

v^2=u^2+2as

v^2=0^2+2\times 1.479\times 140

v^2=414.12

v = 20.349 m/sec

So speed of the car when it passed 0.14 km is 20.349 m/sec

5 0
3 years ago
A car moves with a speed 72km/h, the driver uses the brakes, the car stops after 8 seconds, calculate its speed after 10 seconds
vampirchik [111]
The acceleration of the car is solved by subtracting the initial speed from the final speed then dividing the result by the elapsed time.

initial speed = 72 km/hr = 20 m/s

final speed = 0 m/s

elapsed time = 5 seconds

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acceleration = - 20m/s / 5 s

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A box fails to slide down a ramp at a warehouse. what happens if you put the box on a cart that has wheels?
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A spring with a spring constant of 120 J/m2 is fixed to a wall, free to oscillate. On the other end, a ball with a mass of 1500
Neporo4naja [7]

Answer:

A. 4.47 m/s

Explanation:

As the ball oscillates, it mechanical energy, aka the total kinetic and elastics energy stays the same. For the ball to be at maximum speed, its elastic energy i 0 and vice versa. When the ball is at rest, its kinetic energy is 0 and its elastic energy is at maximum at 50 cm, or 0.5 m

1500 g = 1.5 kg

E_e = E_k

kx^2/2 = mv^2/2

120*0.5^2/2 = 1.5*v^2/2

15 = 0.75v^2

v^2 = 15 / 0.75 = 20

v = \sqrt{20} = 4.47 m/s

5 0
4 years ago
A 1.10 μF capacitor is connected in series with a 1.92 μF capacitor. The 1.10 μF capacitor carries a charge of +10.1 μC on one p
miss Akunina [59]

Answer:

(a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

Explanation:

Given that,

Charge = 10.1 μC

Capacitor C₁ = 1.10 μF

Capacitor C₂ = 1.92 μF

Capacitor C₃ = 1.10 μF

Potential V₁ = 51.5 V

Let V₁ and V₂ be the potentials on the two plates of the capacitor.

(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor

Using formula of potential difference

V_{1}=\dfrac{Q}{C_{1}}

Put the value into the formula

V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}

V_{1}=9.18\ V

The potential on the second plate

V_{2}=V-V_{1}

V_{2}=51.5 -9.18

V_{2}=42.32\ v

(b). We need to calculate the equivalent capacitance of the two capacitors

Using formula of equivalent capacitance

C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}

Put the value into the formula

C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}

C=6.99\times10^{-7}\ F

C=0.69\ \mu F

Hence, (a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

7 0
3 years ago
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