A small object with mass m , charge q , and initial speed v 0 = 6.00×10 3 m/s is

Question

A small object with mass m, charge q, and initial speed v0 = 6.00×103 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Figure 1) . The electric field between the plates is directed downward and has magnitudeE = 700 N/C . Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distanced = 1.35 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance.

Calculate the object's charge-to-mass ratio, q/m.

Explanation / Answer

Given that :

initial speed of small object, v0 = 6 x 103 m/s

separation distance b/w parallel plates, d = 0.26 m

time taken by small object which is given as :

t = d / v0 = (0.26 m) / (6 x 103 m/s)

t = 4.34 x 10-5 sec

Electric field b/w the plates, E = 700 N/C

using equation of motion 2, we get

xf = x0 + v0 t + (1/2) a t2

(0.56 m) = (0.0135 m) + (6 x 103 m/s) (4.34 x 10-5 s) + (0.5) a (4.34 x 10-5 s)2

(0.5465 m) = (0.2604 m) + (9.4178 x 10-10 s2) a

a = (0.2861 m) / (9.4178 x 10-10 s2)

a = 3.03 x 108 m/s2

The object's charge-to-mass ratio, q/m which will be given as :

we know that,   a = q E / m

q / m = (3.03 x 108 m/s2) / (700 N/C)

q / m = 4.32 x 105 C/kg

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