Question

Problem: A point charge (q = +6.00 nC, m= 1.00 x 10-15 kg) is fired from the negative plate of a parallel plate capacitor with an initial speed of vo = 3.50 x 105 m/s. When the particle has traveled two-thirds the distance to the positive plate, its speed is vf= 2.21 x 105 m/s. If the electric field strength in the capacitor is E = 2.15 x10 V/m, what is the capacitor's plate separation, d? a. Use energy conservation to write an expression for d. Your expression should be simplified and contain only the variable names given in the problem. b. Use your expression from (a) to calculate d. Submit this answer below.

Answer 1

We have charge q=6.0 nC, mass m=10^-15kg.

Decrease in kinetic energy of charge

i -i = m : ( - ) = 3.683 * 10-5/

This decrease in KE must be equal to the gain in Electric potential energy.

Gain in E.P.E = qE(2d/3)-E(0) = 2qEd/3

Thus, using conservation of energy,

$\frac{2qEd}{3}=\frac{1}{2}m\left ( v_{0}^{2}-v_{f}^{2} \right )$

i.e. $d=\frac{3}{4qE}m\left ( v_{0}^{2}-v_{f}^{2} \right )$

For given v₀=3.5*10⁵ m/s, v_f=2.21*10⁵ m/s and E=2.15*10⁶ V/m, q=6nC

we get d=1.711cm