# 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

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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

A rock is thrown from the top of a building 146 m high, with a speed of 14 m/s at an angle 43 degrees above the horizontal. When it hits the ground, what is the magnitude of its velocity (i.e. its speed).

time is 32 s and speed is 304.3 m/s

Explanation:

Height, h = 146 m

speed, u = 14 m/s

Angle, A = 43 degree

Let it hits the ground after time t.

Use second equation of motion

Time cannot be negative so the time is t = 32 s .

The vertical velocity at the time of strike is

v' =  u sin A - g t

v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s

horizontal velocity

v'' = 14 cos 43 =10.3 m/s

The resultant velocity at the time of strike is

A force of 140 140 newtons is required to hold a spring that has been stretched from its natural length of 40 cm to a length of 60 cm. Find the work done in stretching the spring from 60 cm to 65 cm. First, setup an integral and find a a, b b, and f ( x ) f(x) which would compute the amount of work done.

The work done in stretching the spring is 0.875 J.

Explanation:

Given that,

Force = 140 N

Natural length = 60-40 = 20 cm

Stretch length of the spring = 65-60 = 5 cm

We need to calculate the spring constant

Using formula of Hooke's law

We need to calculate the work done

On integration

Hence, The work done in stretching the spring is 0.875 J.

A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads share the 28 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cmapart, the force between them is 4.8×10^−4 N . Assume that A has a greater charge. What is the charge qA and qB on the beads?

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} =

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

OFFERING 60 POINTS IF YOU CAN SHOW THE WORK!!!!A 1000 kg roller coaster begins on a 10 m tall hill with an initial velocity of 6m/s and travels down before traveling up a second hill. As the coaster moves from its initial height to its lowest position, 1700J of energy is transformed to thermal energy by friction.

Explanation:

(a) At the top of the hill, the coaster has total energy (potential and kinetic)

E = (1000 kg) g (10 m) + 1/2 (1000 kg) (6 m/s)² = 116,000 J

As it reaches its lowest position, its potential energy is converted to kinetic energy, and some is lost to friction, making its speed v such that

1/2 (1000 kg) v ² = 116,000 J - 1700 J = 114,300 J

===>   v ≈ 15.2 m/s

If no energy is lost to friction as the coaster makes its way up the second hill, all of its kinetic energy would be converted to potential energy at the maximum possible height H.

1/2 (1000 kg) (15.2 m/s)² = (1000 kg) gH

===>   H11.7 m

(b) At the top of the second hill with minimum height h, and with maximum speed 4.6 m/s, the coaster has energy

E = P + K = (1000 kg) gh + 1/2 (1000 kg) (4.6 m/s)²

Assuming friction isn't a factor again, the energy here should match the energy at the lowest point in part (a), 114,300 J.

(1000 kg) g h + 1/2 (1000 kg) (4.6 m/s)² = 114,300 J

===>   h10.6 m

Tonya picks up a leaf from the ground and holds it at arm’s length. She lets go, and the leaf falls to the ground. What force pulled the leaf to the ground?

Gravity

Explanation:

Gravity is constantly pulling objects downward. Without it, everything would float out into space.

I hope this answer helps :)

The answer for the given question above would be option C. GRAVITATIONAL FORCE. Based on the given scenario above of a leaf that falls to the ground when Tonya let it go, the force that pulled the leaf to the ground is the gravitational force. This kind of force is a force that attracts any object with mass.

Hope this helps!!!

wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength

Complete question:

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?

b) the center of the left ring?

a) the electric field strength at the midpoint between the two rings is 0

b) the electric field strength at the center of the left ring is 2712.44 N/C

Explanation:

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r =

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

The electric field strength at the midpoint;

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.