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Greeley [361]
2 years ago
11

A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates. A 1.0 g plastic bead, with a charg

e of -6.0 nC, is suspended between the two plates by the force of the electric field between them.
Required:
a. Which plate, the upper or the lower, is positively charged?
b. What is the charges on positvie plate?
Physics
1 answer:
marissa [1.9K]2 years ago
3 0

Answer:

Please find the answer in the explanation

Explanation:

Given that A 1.0 g plastic bead, with a charge of -6.0 nC, is suspended between the two plates by the force of the electric field between them.

Since it is suspended, it must have been repelled by the bottom negative plate and trying to be attracted to the top plate.

We can therefore conclude that the upper plate, is positively charged

B.) The charge on the positive plate of parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates must be less than 6.0 nC

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A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T
atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = \rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.

The magnitude of the torque on the dipole = \rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.

<em>where</em>,

k is the Coulomb's constant, having value = \dfrac{1}{4\pi \epsilon_o}

\epsilon_o is the electrical permittivity of free space.

Also, r>>s, therefore, \rm r^2+s^2\approx r^2.

\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

\rm F=qE

Therefore, the electric force on the charge Q due to the dipole is given by

\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.

The magnitude of the torque on the dipole is given by

\rm \tau = Fs\ \sin\theta

Since the charge Q is placed in the plane that bisects the dipole, therefore, \theta = 90^\circ.

\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.

4 0
2 years ago
A plane increases its speed from 450 km/hr to 750 km/hr in 0.30 hour. What is its acceleration?
ryzh [129]

Answer:

The acceleration is found as:

a = 1000 km/h²

Explanation:

Initial speed of the plane = 450 km/h

Final speed of the plane = 750 km/h

Time taken = 0.3 h

Acceleration can be defined as the change of speed of the object divided by the time taken to bring that change

Acceleration is given as:

a=\frac{v_f-v_i}{t}

where v(f) = 750 km/h , v(i) = 450 km/h , t = 0.3 h

Substitute the values in the equation of acceleration

a= \frac{750-450}{0.3}\\a=\frac{300}{0.3}\\a=1000 km/h^2

This is the found result

5 0
2 years ago
The same physics student jumps off the back of her Laser again, but this time the Laser is
soldi70 [24.7K]

a) The speed of the student after the jump is 1.07 m/s

b) The final speed of the laser is 10.4 m/s

Explanation:

a)

We can solve this problem by applying the law of conservation of momentum: if there are no external forces acting on the system, the total momentum of the student+Laser system must be constant. Therefore, we can write:

p_i = p_f\\0=mv+MV

where

The initial momentum is zero

m = 42 kg is the mass of the Laser

v = 1.5 m/s is the final velocity of the Laser

M = 59 kg is the mass of the student

V is the final velocity of the student

Solving the equation for V, we find the velocity of the student:

V=-\frac{mv}{M}=-\frac{(42)(1.5)}{59}=-1.07 m/s

So, the final speed of the student is 1.07 m/s.

b)

In this case, the laser and the student are travelling at 3.1 m/s before the student jumps off: therefore, the total momentum before the jump is not zero.

So, the equation of the conservation of momentum is

(m+M)u=mv+MV

where

m = 42 kg is the mass of the Laser

M = 59 kg is the student's mass

u = 3.1 m/s is the initial velocity of the student and the Laser

V = -2.1 m/s is the velocity of the student after the jump (she jumps backward)

v is the final velocity of the Laser

And solving for v, we find

v=\frac{(m+M)u-MV}{m}=\frac{(42+59)(3.1)-(59)(-2.1)}{42}=10.4 m/s

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

3 0
2 years ago
Read 2 more answers
A light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N
Mnenie [13.5K]

Answer:

8N and 32N

Explanation:

Given that a  light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N weight is placed between the two sawhorses, 3.0 m from the edge and 2.0 m from the center.

To calculate the forces that are exerted by the sawhorses on the board, we must consider the equilibrium of forces acting on the board.

Let the two upward forces produce by the saw horses be P1 and P2

Assuming that the weight is negligible

Sum of the upward forces = sum of the downward forces.

P1 + P2 = 40 ....... (1)

Also, the sum of the clockwise moment = sum of the anticlockwise moments.

Let's assume that the board is uniform. The weight will act at the centre.

Taking moment at the centre:

P1 × 5 + 40 × 2 = 0

P1 = 40 / 5

P1 = 8N

Substitute P1 into equation 1

8 + P2 = 40

P2 = 40 - 8

P2 = 32N

3 0
2 years ago
Read 2 more answers
An object is released from rest near and above Earth’s surface from a distance of 10m. After applying the appropriate kinematic
Damm [24]

Answer:

v_y = 12.54 m/s

Explanation:

Given:

- Initial vertical distance y_o = 10 m

- Initial velocity v_y,o  = 0 m/s

- The acceleration of object in air = a_y

- The actual time taken to reach ground t = 3.2 s

Find:

- Determine the actual speed of the object when it reaches the ground?

Solution:

- Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:

                             y = y_o + v_y,o*t + 0.5*a_y*t^2

                             0 = 10 + 0 + 0.5*a_y*(3.2)^2

                             a_y = - 20 / (3.2)^2 = 1.953125 m/s^2

- Use the principle of conservation of total energy of system:

                             E_p - W_f = E_k

Where,                  E_p = m*g*y_o

                             W_f = m*a_y*(y_i - y_f)      ..... Effects of air resistance

                             E_k = 0.5*m*v_y^2

Hence,                  m*g*y_o - m*a_y*(y_i - y_f) = 0.5*m*v_y^2

                             g*(10) - (1.953125)*(10) = 0.5*v_y^2

                             v_y = sqrt (157.1375)

                            v_y = 12.54 m/s

4 0
2 years ago
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