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Anna35 [415]
4 years ago
9

Two large, parallel, conducting plates are 21 cm apart and have charges of equal magnitude and opposite sign on their facing sur

faces. An electrostatic force of 3.2 10-15 N acts on an electron placed anywhere between the two plates.
(Neglect fringing).



(a) Find the magnitude of the electric field at the position of the electron.
______ N/C



(b) What is the potential difference between the plates?
______ V
Physics
1 answer:
Tresset [83]4 years ago
4 0

Explanation:

It is given that,

Distance between two conducting plates, d = 21 cm = 0.21 m

Electrostatic force acting on the electron placed anywhere between the two plates, F=3.2\times 10^{-15}\ N

(a) Let E is the electric field at the position of the electron. The electric force acting on a charged particle per unit charge is called electric field. It is given by :

E=\dfrac{F}{q}

q is the charge on electron

E=\dfrac{3.2\times 10^{-15}\ N}{1.6\times 10^{-19}\ C}

E = 20000 N

(b) The V is the potential difference between the plates. The relation between the potential difference and the electric field is given by :

V=E\times d

V=20000\ N\times 0.21\ m

V = 4200 volts

Hence, this is the required solution.

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A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is
lys-0071 [83]

Answer:

\omega_f=571.42\ rpm

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, a=1.5\ m/s^2

Displacement, d = 1.3 m

The angular acceleration is given by :

\alpha =\dfrac{a}{r}

\alpha =\dfrac{1.5}{0.033}

\alpha =45.46\ rad/s^2

The angular displacement is given by :

\theta=\dfrac{d}{r}

\theta=\dfrac{1.3}{0.033}

\theta=39.39\ rad

Using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

Here, \omega_i=0

\omega_f=\sqrt{2\alpha \theta}

\omega_f=\sqrt{2\times 45.46\times 39.39}

\omega_f=59.84\ rad/s

Since, 1 rad/s = 9.54 rpm

So,

\omega_f=571.42\ rpm

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0
4 years ago
Timmy drove 2/5 of a journey at an average speed of 20 mph.
mixer [17]

Answer:

4hr

Explanation:

5 0
3 years ago
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What is the mass of an object that weighs 686N on Earth?
Oxana [17]

Answer:

70 kg is the mass of the object

Explanation:

This question can be solved with this simple formula:

Weight force = mass . gravity

686 N = mass . 9.8 m/s²

686 N /  9.8 m/s² = mass → 70 kg

Note → 1N = 1 kg . m / s²

8 0
4 years ago
A circular loop with radius r is rotating with constant angular velocity ω in a uniform electric field with magnitude E. The axi
inn [45]

Answer:

\Phi_{E} = E\pi r^2 \omega t

Explanation:

The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

\Phi_{E} = \vec{E}\vec{A}

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.

The above formula can be rewritten as follows

\Phi_{E} = EA\cos(\theta)

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.

If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

\theta = \omega t

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.

Therefore the electric flux can be written as a function of time

\Phi_{E} = E\pi r^2 \omega t

3 0
4 years ago
Why are we not crushed by the weight of the atmosphere on our shoulders?
antiseptic1488 [7]

Answer:

Due to equal pressure in all the direction at a particular level in a fluid medium (Pascal's Law)

Explanation:

We are not crushed by the weight of the atmosphere because atmosphere is a fluid and we are immersed into it. So, according to the Pascal's law the the pressure a each point in a horizontal level is equal in all the direction irrespective of the orientation of a body.

Variation of pressure in term of the height of a fluid medium is given as:

P=\rho.g.h

\rho=density of fluid

g = acceleration due to gravity

h = height of the free surface of the fluid from the immersed object.

  • And atmosphere has very less variation of pressure with change in height as it is a rare medium fluid and so for a human height there is very negligible variation of pressure at the heat of a human with respect to his toe.

5 0
4 years ago
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