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= 1.07 x 105 m/s. If the electric field strength in the capacitor is E = 2.27x106 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

a ) .Here the mechanical energy of charge q will be constant as the force acting on charge q is conservative .

Therefore ,

-Change in kinetic energy = change in potential energy

Change in kinetic energy = $\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{0}^{2}$

Potential energy at a point is equal to charge times potential at that point.Therefore, change in potential energy will be equal to charge times potential difference .Potential difference is given by Ed(d is the separation between the two points) in the case of parallel plate capacitor ..

Here , the distance travelled by the charge q is 2/3 of the separation between the plates i.e 2d/3 .

Thus , potential difference will be E*2d/3 .

Change in potential energy will be q*E*2d/3 i.e 2qEd/3 .

Now ,

-Change in kinetic energy = change in potential energy

$-\left ( \frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{0}^{2}\right )=\frac{2*q*E*d}{3}$

$\Rightarrow \frac{1}{2}m(v_{0}^{2}-v_{f}^{2})=\frac{2*q*E*d}{3}$

$\Rightarrow \frac{3m}{4qE}(v_{0}^{2}-v_{f}^{2})=d$

b ) .

m = 1*10^-15 kg

q = +6nC = 6 * 10^-9 C

E = 2.27*10⁶ V/m

v_f = 1.07 * 10⁵ m/s

v₀ = 3.5*10⁵ m/s

$d=\frac{3*10^{-15}(3.5^{2}*10^{10}-1.07^{2}*10^{10})}{4*6*10^{-9}*2.27*10^{6}}\: meter$

d = 6.12 cm