At a given instant, a 2.2 A current flows in the wires connected to a parallel-plate capacitor. What is the rate at which the electric field is changing between the plates if the square plates are 2.0 cm on a side? Express your answer using two significant figures

Answers

Answer 1
Answer:

Answer:

Check attachment for better understanding

Explanation:

Given that,

Current in wire I =2.2A

Capacitor plate dimension is 2cm by 2cm

s=2cm=2/100 = 0.02m

Rate at which electric field Is changing dE/dt?

The current in the wires must also be the displacement current in the capacitor. We find the rate at which the electric field is changing from

ID = ε0•A•dE/dt

Where ε0 is a constant

ε0= 8.85×10^-12C²/Nm²

Area of the square plate is

A =s² =0.02² = 0.0004m²

Then,

Make dE/dt the subject of formula

dE/dt = ID/ε0A

dE/dt = 2.2 / (8.85×10^-12 ×4×10^-4)

dE/dt = 6.215×10^14 V/m-s

Or

dE/dt = 6.215×10^14 N/C.s

The rate at which the electric field is changing between the plates is 6.215×10^14 N/C.s


Related Questions

An advantage of J.J. Thomson's Plum Pudding Model was that it _____. A. was a much less expensive way to study atoms B. simplified the calculations necessary to describe an atomC. clearly explained where electrons were located in an atomD. is much less expensive to bake a plum pudding than to look at an atom
What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?
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A. With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 45m?b. How long will it be in the air?

What happens to the pressure in all parts of a confined fluid if the pressure in one part is increased? The pressure in the other parts remains the same.The pressure everywhere increases by different amounts depending on the area of each part.
The pressure everywhere increases by the same amount.
The pressure everywhere decreases to conserve total pressure.

Answers

Answer:

option C

Explanation:

the correct answer is option C

When in a confined fluid the pressure is increased in one part than the pressure will equally distribute in the whole system.

According to Pascal's law when pressure is increased in the confined system then the pressure will equally transfer in the whole system.  

This law's application is used in machines like hydraulic jacks.

In a game of tug of war, a rope is pulled by a force of 182 N to the right and by a force of 108 N to the left. Calculate the magnitude and direction of the net horizontal force on the rope.

Answers

In a game of tug of war the magnitude and direction of the net horizontal force on the rope is 74 N.

What is force?

A force in physics is an effect that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

In a game of tug of war, a rope is pulled by a force of 182 N to the right and by a force of 108 N to the left. The magnitude and direction of the net horizontal force on the rope is,

F = 182 - 108

F = 74 N

The magnitude and direction of the net horizontal force on the rope is  74 N.

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Answer:

74 N to the right

Explanation:

the forces are going in opposite horizontal directions, meaning that they are directly opposing each other. this means that you can subtract the force applied in the direction that is greater from the direction that is less to get the net force for the greater direction

this means 182 N - 108 N = 74 N to the right

Determine the angular velocity of the merry-go-round if a jumps off horizontally in the −n direction with a speed of 2 m/s , measured relative to the merry-go-round. neglect friction and the size of each child.

Answers

by angular momentum conservation we will have

angular momentum of child + angular momentum of merry go round = 0

angular momentum of child = mvR

m = mass of child

R = radius of child

v = speed = 2 m/s

now let's say moment of inertia of merry go round is I

so we will have

m*2*R + Iw = 0

w = -(2mR)/(I)

so merry go round will turn in opposite direction with above speed

Suppose that the speedometer of a truck is set to read the linear speed of the truck, but uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the truck, will the reading on the speedometer be correct? If not, will the reading be greater than or less than the true linear speed of the truck? Why?

Answers

The measurement will be significantly affected.

Recall that the relationship between linear velocity and angular velocity is subject to the formula

v = \omega r,

Where r indicates the radius and \omega the angular velocity.

As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.

How many nanoseconds are in one hour? How do you write the following in scientific notation?2,560,000m

Answers

Answer:

3.6 × 10¹² nanoseconds

Explanation:

Hour is the unit of time. Seconds is the SI unit of time.

Hour and seconds are related as:

1 hour = 60 minutes

1 minute = 60 seconds

So,

1 hour = 60 ×60 seconds = 3600 seconds

Thus,

3600  seconds are in one hour

Also,

1 sec = 10⁹ nanoseconds

Thus,

3600 sec = 3600 × 10⁹ nanoseconds = 3.6 × 10¹² nanoseconds

Thus,

3.6 × 10¹² nanoseconds are in one hour.

An infant throws 7 g of applesauce at a velocity of 0.5 m/s. All of the applesauce collides with a nearby wall and sticks to it. What is the decrease in kinetic energy of the applesauce?

Answers

Answer:

Δ KE = - 8.75 x 10⁻⁴ J

Explanation:

given,

mass of applesauce = 7 g = 0.007 Kg

initial velocity, u = 0.5 m/s

final velocity, v = 0 m/s

Decrease in kinetic energy = ?

initial kinetic energy

KE_1=(1)/(2)mu^2

KE_1=(1)/(2)* 0.007 * 0.5^2

      KE₁ = 8.75 x 10⁻⁴ J

final kinetic energy

KE_2=(1)/(2)mv^2

KE_2=(1)/(2)* 0.007 * 0^2

      KE₂ =0 J

Decrease in kinetic energy

Δ KE =  KE₂ - KE₁

Δ KE = 0 - 8.75 x 10⁻⁴

Δ KE = - 8.75 x 10⁻⁴ J

decrease in kinetic energy of the applesauce is equal to  8.75 x 10⁻⁴ J

Final answer:

The decrease in kinetic energy of the applesauce, when it hits the wall and stops, is the initial kinetic energy of it. Using the formula of kinetic energy, the decrease is calculated to be 0.000875 Joules.

Explanation:

This question relates to the concept of kinetic energy in physics. Kinetic energy is calculated by the formula 0.5 * mass (kg) * velocity (m/s)^2. So the initial kinetic energy of the applesauce right after being thrown was 0.5 * 0.007 kg * (0.5 m/s)^2 = 0.000875 Joules.

When the applesauce hits the wall and stops, its velocity drops to 0. Thus, its kinetic energy also goes to 0 (because kinetic energy is proportional to the square of velocity).

Therefore, the decrease in kinetic energy is the same as the initial kinetic energy of the applesauce, which is 0.000875 Joules.

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