PHYSICS COLLEGE

The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.a. What is the electric field strength and direction between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?

Answers

Answer 1
Answer:

Answer:

A. 2.083 MV/m from anode to cathode.

B. 93648278.15 m/s

C. 2.5x10^-5 C and there are about 1.56x10^14 electrons

D. 4x10^-15 Joules

Explanation:

Voltage V across plate is 25 kV = 25x10^3 V

Distance apart x = 1.2 cm = 1.2x10^-2 m

A. Electric field strength is the potential difference per unit distance

E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m

= 2.083 MV/m

B. Energy of electron is electron charge times the voltage across

i.e eV

Charge on electron = 1.6x10^-19 C

Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules

Mass of electron m is 9.12x10^-31 kg

Kinetic energy of electron = 0.5mv^2

Where v is the speed

4x10^-15 = 0.5 x 9.12x10^-31 x v^2

v^2 = 8.77x10^15

v = 93648278.15 m/s

C. From Q = CV

Q = charge

C = capacitance = 1 nF 1x10^-9 F

V = voltage = 25x10^3 V

Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C

Total number of electrons = Q/e

= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons

D. To push electron from cathode to anode, I'll have to do a work of about

4x10^-15 Joules

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The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally from west to east. the vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of 3.00 10-5 t. (a) in what direction are the electrons deflected by this field component? due north due south due east due west (b) what is the magnitude of the acceleration of an electron in part (a)? m/s2

Answers

(a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
- index finger: velocity --> due east
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However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
K= (1)/(2)mv^2
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
v= \sqrt{ (2K)/(m) }= \sqrt{ (2 \cdot 2.20 \cdot 10^(-15) J)/(9.1 \cdot 10^(-31) kg) }=6.95 \cdot 10^7 m/s

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
qvB = m (v^2)/(r)
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
r= (mv)/(qB)= ((9.1 \cdot 10^(-31) kg)(6.95 \cdot 10^7 m/s))/((1.6 \cdot 10^(-19) C)(3.00 \cdot 10^(-5) T))=13.18 m

And finally we can calculate the centripetal acceleration, given by:
a_c = (v^2)/(r)= ((6.95 \cdot 10^7 m/s)^2)/(13.18 m)=3.66 \cdot 10^(14) m/s^2

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Answers

Answer:

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Explanation:

At what distance from a long straight wire carrying acurrentof 5.0A is the magnitude of the magnetic field due to the
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?

Answers

Answer:

The distance is 2 cm

Solution:

According to the question:

Magnetic field of Earth, B_{E} = 5.0* 10^(- 5) T

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

B = \farc{\mu_(o)I}{2\pi d}

where

d = distance from current carrying wire

Now,

d = (\mu_(o)I)/(2\pi B)

d = (4\pi* 10^(- 7)* 5.0)/(2\pi* 5.0* 10^(- 5))

d = 0.02 m 2 cm

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