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Artemon [7]
2 years ago
5

A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th

e surface. Part A) What is the strength of the electric field 0.1 mm above the center of the top surface of the plate? Express your answer to two significant figures and include the appropriate units. Part B) What is the strength of the electric field 0.1 mm below the center of the bottom surface of the plate? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
son4ous [18]2 years ago
5 0

Answer:

Part a)

E = 8436.7 N/C

Part b)

E = 8436.7 N/C

Explanation:

Part a)

Electric field due to large sheet is given as

E = \frac{\sigma}{2\epsilon_0}

\sigma = \frac{Q}{A}

Q = -3.8 nC

A = \pi(0.09)^2

A = 0.025 m^2

\sigma = \frac{-3.8\times 10^{-9}}{0.025}

\sigma = -1.5 \times 10^{-7} C/m^2

now the electric field is given as

E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}

E = 8436.7 N/C

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

E = 8436.7 N/C

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A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18
nevsk [136]

Answer:

(I). The motional emf induced between the ends of the segment is 2.88i-0.72k

(II). The motional emf is zero.

Explanation:

Given that,

Magnetic field = 0.080 T

Velocity of wire segment = 78 m/s

Component in x direction = 18 m/s

Component in y direction = 24 m/s

Component in z direction = 72 m/s

Length = 0.50 m

We need to calculate the motional emf induced between the ends of the segment

Using formula of emf

\epsilon=(B_{x}\times v_{x})l

Put the value into he formula

\epsilon=((0,0.080,0)\times(18,24,72))\times0.50

\epsilon=((72\times0.080)i-0j+k(-18\times0.080))\times0.50

\epsilon=(5.76i-0j-1.44k)\times0.05

\epsilon=2.88i-0.72k

(II). If the wire segment is parallel to the y-axis then the angle between B and v is zero.

We need to calculate the motional emf

Using formula of emf

\epsilon =Bv\sin\theta\times l

Here, \theta = 0

\epsilon =Bv\sin0\times l

\epsilon = 0

Hence, (I). The motional emf induced between the ends of the segment is 2.88i-0.72k

(II). The motional emf is zero.

6 0
2 years ago
If you wanted to measure the width of the gym, how would the accuracy of a meter stick compare with that of a 50m tape
Alisiya [41]
Meter stick would not be as accurate,
Every time you placed it down and picked it back up you run the chance of losing 2-4 cm each time.
7 0
2 years ago
Suggest reasons why poaching for subsistence is likely to be less damaging to the biodiversity of an area than poaching for prof
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5 0
2 years ago
A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T
DerKrebs [107]

Answer:

0.087976 Nm

Explanation:

The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;

τ = NIAB sinθ     --------- (i)

Where;

N = number of turns of the loop

I = current in the loop

A = area of each of the turns

B = magnetic field

θ = angle the loop makes with the magnetic field

<em>From the question;</em>

N = 200

I = 4.0A

B = 0.70T

θ = 30°

A = π d² / 4        [d = diameter of the coil = 2.0cm = 0.02m]

A = π x 0.02² / 4 = 0.0003142m²         [taking π = 3.142]

<em>Substitute these values into equation (i) as follows;</em>

τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°

τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5

τ = 200 x 4.0 x 0.0003142 x 0.70      

τ = 0.087976 Nm

Therefore, the torque on the coil is 0.087976 Nm

3 0
2 years ago
Which of the following statements are true about an ideal solution of two volatile liquids? A. The partial pressure of each comp
Zanzabum

the correct choices are

A. The partial pressure of each component above the liquid is given by Raoult's law

and

C. An ideal solution of two volatile liquids can exist over a range of pressures that are limited by the pressure for which only a trace of liquid remains, and the pressure for which only a trace of gas remains

in ideal solution , when two volatile liquids are mixed no energy change takes place in the energy of the solution.


4 0
2 years ago
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