An electron starts from one plate of a charged closely spaced (vertical) paralle

Question

An electron starts from one plate of a charged closely spaced (vertical) parallel plate arrangement with a velocity of 1.16×104 m/s to the right. Its speed on reaching the other plate, 1.95 cm away, is 4.71×104 m/s .

Part A

What type of charge is on each plate?

Part B

What is the direction of the electric field between the plates?

Part C

If the plates are square with an edge length of 24.0 cm , determine the charge on each. Answer in C

Thank You!

A. positive on right plate and negative on left plate B. negative on right plate and positive on left plate

Explanation / Answer

Here ,

A)

as the final speed of electron is more than it's initial speed ,

hence , the plate on the right is positive plate

the correct choice is

A.   positive on right plate and negative on left plate.

B)

force on electron is in the direction oposite to the applied field.

as elctron is accelerating towards right

the direction of electric field is

A) from right to left

C)

v = 4.71 *10^4 m/s

u = 1.16 *10^4 m/s

let acceleration of electon is a

v^2 - u^2 = 2 * a* d

(4.71 *10^4)^2 - (1.16 *10^4)^2 = 2 * a * 0.0195

a = 5.343 *10^10 m/s^2

Now ,

let charge on plates is q

as the electric field E = q/(epsilon * area)

q/(epsilon * area) = mass * a /e

q = 9.1 *10^-31 * 5.343 *10^10 * 0.24^2 * 8.854 *10^-12 /(1.602 *10^-19 )

q = 1.55 *10^-13 C

th charge on each is 1.55 *10^-13 C

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