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Suppose an electron is released from rest in a uniform electric field whose magnitude is 5.50 103 V/m.

a) Through what potential difference will it have passed after moving 3.95 cm?
b) How fast will the electron be moving after it has traveled 3.95 cm?

User Maikon
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1 Answer

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Final answer:

a) The potential difference will be 217.25 V. b) The electron will be moving at a velocity of approximately -7.91 x 10^4 m/s.

Step-by-step explanation:

To solve this question, we need to use the formula V = Ed, where V is the potential difference, E is the electric field strength, and d is the distance.

a)

Plugging in the given values: E = 5.50 x 103 V/m and d = 3.95 cm = 0.0395 m, we can calculate the potential difference V.

V = (5.50 x 103 V/m) x (0.0395 m) = 217.25 V

b)

Using the formula v = at, where v is the final velocity, a is the acceleration, and t is the time, we can calculate the final velocity of the electron.

Since the electron starts from rest, we know that the initial velocity (u) is 0 m/s. We also know that the acceleration (a) is given by a = qE/m, where q is the charge of the electron and m is its mass.

Plugging in the given values: q = -1.60 x 10-19 C and m = 9.11 x 10-31 kg, we can calculate the acceleration a.

a = (-1.60 x 10-19 C x 5.50 x 103 V/m) / (9.11 x 10-31 kg) ≈ -9.17 x 1010 m/s²

Now we can use the formula v = at, where v is the final velocity, a is the acceleration, and t is the time. Since the electron travels a distance of 0.0395 m, we can calculate the time t.

0.0395 m = 0 x t + 0.5 x (-9.17 x 1010 m/s²) x t²

Simplifying the equation, we get:

-4.585 x 109 t² = 0.0395

t² = 0.0395 / -4.585 x 109

t ≈ √(0.0395 / -4.585 x 109) = 8.63 x 10-6 s

Finally, we can calculate the final velocity using the formula v = at.

v = (-9.17 x 1010 m/s²) x (8.63 x 10-6 s) ≈ -7.91 x 104 m/s

User Ajitspyd
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