A small object with mass m , charge q , and initial speed v 0 = 4.00×10 3 m/s is

Question

A small object with mass m, charge q, and initial speed v0 = 4.00×103 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm. The electric field between the plates is directed downward and has magnitude E = 600 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 = 1.35 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

let the acceleration due to electric field is a

time taken for the object to cross plates

t1 = 0.26/(4 *10^3) s = 6.5 *10^-5 s

Now, vertical velocity just after exiting electric field region

vy = a * t1 = 6.5 *10^-5 *a

Now, time taken to reach the cross the area with no electric field

t2 = 0.56/(4 *10^3) = 1.4 *10^-4 s

Now, for the distance d

0.0135 = 0.5 * a * (6.5 *10^-5)^2 + (a* 6.5 *10^-5) * 1.4 *10^-4

solving for a

a = 1.204 *10^6 m/s^2

Now, using second law of motion

m * a = q * E
1.204 *10^6 * m = q * 600

solving

q/m = 2007 C/Kg

the the object's charge-to-mass ratio is 2007 C/Kg

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