At some instant the velocity components of an electron moving between two charged parallel plates are vx = 1.3 105 m/s and vy = 3.6 103 m/s. Suppose that the electric field between the plates is given by E = (120 N/C) j.

(a) What is the electron's acceleration in the field?
m/s2 j
(b) What is the electron's velocity when its x coordinate has changed by 4.8 cm?
m/s i + m/s j

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To answer these questions, we need to use the principles of motion in an electric field. Let's break down the problem step by step.

(a) To find the electron's acceleration in the field, we can use the equation:

a = q/m * E

Where a is the acceleration, q is the charge of the electron, m is the mass of the electron, and E is the electric field.

The charge of an electron is q = -1.6 x 10^-19 C, and the mass of an electron is m = 9.1 x 10^-31 kg. Given that the electric field E = (120 N/C)j, we can substitute these values into the equation:

a = (-1.6 x 10^-19 C) / (9.1 x 10^-31 kg) * (120 N/C) j

Simplifying the calculation, we get:

a = -2.09 x 10^11 m/s^2 j

Therefore, the electron's acceleration in the field is -2.09 x 10^11 m/s^2 in the j direction.

(b) To find the electron's velocity when its x coordinate has changed by 4.8 cm, we can use the equation:

v = vx + a * t

Where v is the velocity, vx is the initial x-component of velocity, a is the acceleration, and t is the time.

Since the only acceleration is in the j direction, we can ignore the acceleration component in the x direction. Therefore, the equation becomes:

v = vx

Given that vx = 1.3 x 10^5 m/s, the electron's velocity when its x coordinate has changed by 4.8 cm is:

v = (1.3 x 10^5 m/s)i + 0 j

Therefore, the electron's velocity is (1.3 x 10^5 m/s)i + 0j.