An electron is moving to the right with an initial speed of 2.40 x 10^6 m/s, whe

Question

An electron is moving to the right with an initial speed of 2.40 x 10^6 m/s, when it enters a region of uniform electric field that points upward and has a magnitude of 1670 N/C. Find the magnitude and direction of the electric force on the electron when It Is in the electric field. Find the magnitude and direction of the electron's acceleration when it is in the electric field. After the electron has been in the electric field for 2.25 ns, what are the x and y component of its velocity? What is its speed?

Explanation / Answer

Here,

initial speed , u = 2.4 *10^6 m/s

electric field , E = 1670 N/C

a) electic force = charge * electric field

electic force = 1.602 *10^-19 * 1670 N in downwards

electic force = 2.68 *10^-16 N in downwards direction

b) Now , as direction of acceleration is in the direction of electric force

acceleration = 2.68 *10^-16/(9.11 *10^-31)

acceleration = 2.94 *10^14 m/s^2 in downwards direction

c)

after 2.25 ns

x component of velocity = 2.4 *10^6 m/s

y component of velocity = a * t

y component of velocity = 2.25 *10^-9 * 2.94 *10^14

y component of velocity = 6.61 *10^5 m/s

speed of electron = sqrt(2.4^2 + 0.661^2) *10^6

speed of electron = 2.49 *10^6 m/s

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