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2 years ago

11

Consider a vacuum-filled parallel plate capacitor (no dielectric material between the plates). What is the ratio of conduction c

urrent Jc to displacement current Jd at 10 MHz

1 answer:

Lorico [155]2 years ago

8 0

Answer: 1798

Explanation:

Given that there is no dielectric material between the plates,

the permittivity of free space = 9 × 10^-12 f/m

The frequency F = 10 MHz

The ratio of conduction current JC to the displacement current Jd is also known as loss tangent.

Please find the attached file for the solution

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A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

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1 year ago

A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.Ex

amid [387]

Answer:

E = k*Q₁/R₁² V/m

V = k*Q₁/R₁ Volt

Explanation:

Given:

- Charge distributed on the sphere is Q₁

- The radius of sphere is R₁

- The electric potential at infinity is 0

Find:

What is the electric field at the surface of the sphere?E.

What is the electric potential at the surface of the sphere?V

Solution:

- The 3 dimensional space around a charge(source) in which its effects is felt is known in the electric field.

- The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.

- If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

F = k*Q₁/R₁²

- Then the electric field at that point is

E = F/1

E = k*Q₁/R₁² V/m

- The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

- Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

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A block of mass 0.1 kg is attached to a spring of spring constant 22 N/m on a frictionless track. The block moves in simple harm

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Answer:

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Explanation:

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On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s

xenn [34]

The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.

First of all, we need to calculate the speed of sound at temperature of $T=31^{\circ}C$ :
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The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
$v= \frac{2L}{t}$
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
$L= \frac{vt}{2}= \frac{(349.6 m/s)/4.60 s)}{2}= 804.1 m$

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2 years ago