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 magnitude E = 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 distance d = 1.15 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

Explanation / Answer

time spent inside the plates t = 26 x 10-2 / vo = 0.26 / 6000 = 4.333333333 x 10-5 seconds

For this much time the charge has been accelerated downwards by the Electric field.

downward acceleration a = Electric force / mass = Eq/m = 700 q/m

vertical distance travelled in time t, s = 0.5 at2 = 0.5 x 700 q/m x ( 4.333333333 x 10-5 )2 = s = 6.572222222 x 10-7 q/m

vertical velocity (downwards) after crossing the plates is v = at = (700q/m) x 4.333333333 x 10-5

v = 0.030333333 q/m

for the rest of the journey, time taken be T = 0.56 / 6000 = 9.333333333 x 10-5 seconds

downward distance further travelled in the second part of the journey s' = vT

s' = (0.030333333 q/m) x9.333333333 x 10-5 = 2.831111111 x 10-6 q/m

total vertical distance travelled d = s + s' = 6.572222222 x 10-7 q/m + 2.831111111 x 10-6 q/m

d = 1.15 cm = 0.0115 m

so, 0.0115 =  6.572222222 x 10-7 q/m + 2.831111111 x 10-6 q/m

0.0115 = (6.572222222 x 10-7 + 2.831111111 x 10-6 )q/m

0.0115 = 3.488333333 x 10-6 q/m

q/m = 3296.703297 C/kg

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