Problem 7: Consider the parallel-plate capacitor shown in the figure. The plate

Question

Problem 7: Consider the parallel-plate capacitor shown in the figure. The plate separation is 3.3 mm and the the electric field inside is 29 N/C. An electron is positioned halfway between the plates and is given some initial velocity, v Randomized Variables d=3.3 mm E = 29 N/C Otheexpertta.com Part (a) What speed, in meters per second, must the electron have in order to make it to the negatively charged plate? Numeric A numeric value is expected and not an expression. The Expert TA I Human-like Grading, Automated Part (b) If the clectron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What will its speed be, in meters per second, when it reaches the positive plate in this case? Numeric : A numeric value is expected and not an expression. vrb 0/15/2017

Explanation / Answer

a)

E = electric field = 29 N/C

d = saperation between the plates = 3.3 mm = 0.0033 m

q = charge on electron = 1.6 x 10-19 C

V = potential difference from halfway to negative plate = E *(d/2) = (29) (0.0033/2)

U = electric potential energy from halfway to left plate = q V

m = mass of electron = 9.1 x 10-31 kg

Vi = speed

Using conservation of energy

Kinetic energy = Electric potential energy

(0.5) m Vi2 = q V

(0.5) (9.1 x 10-31 ) Vi2 = (1.6 x 10-19) (29) (0.0033/2)

Vi = 1.297 x 105 m/s

b)

a = acceleration = qE/m

                         = (1.6 x 10-19) (29) / (9.1 x 10-31)

                         = 5.1 x 1012 m/s2

Vo = initial velocity towards the positive plate after turning around = (1.24 x 105 )/2 = 0.648 x 105

X = displacement = d/2 = 1.65 x 10-3 m

Vf = final velocity at the positive plate

using the equatioin

Vf2 = Vo2 + 2 a X

Vf2 = (0.648 x 105 )2 + 2 (5.1 x 1012) (1.65 x 10-3)

Vf = 1.45 x 105 m/s

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