Question

An electron is released from rest at the negative plate of a parnill plste copcitor. The charge per unit area on cach plate is 1.61 x 10 C/m, and the plate seporation is 1.37 10 m. How fesat is the ectron moving just before it reaches the positive plate? m/s Secton 188 SP Asmall plastic ball with a mass o 6.)ー103kg and with a charge e/.0.136pCissung nd dhan anino dating thread and hangs between the plandacapator (we the de ing). The balbinomhm with the thread making an angle of 30.0 with respect to the vertical. The area of each plate is 0.019 m What is the magnitude of the charge on each plate? Additional Materials

Answer 1

Answer:

(1) Capacitance C = q/V or V = q/C

For a parallel plate capacitor C = $\varepsilon$₀A/d

Therefore, V = q/( $\varepsilon$₀A/d) = d$\sigma$/ $\varepsilon$₀   since $\sigma$ = q/A

Then the electric potential energy U = qV and this must be equal to the kinetic energy of the electron.

that means, KE = U or 1/2mv² = qV = qd$\sigma$/ $\varepsilon$₀

Now, we need to calculate the velocity of the electron just before it reaches the positive plate, therefore,

v² = (2qd$\sigma$/ $\varepsilon$₀) / m = 2qd$\sigma$/ m$\varepsilon$₀

Given that, $\sigma$ = 1.61 x 10^-7C/m², d = 1.37 x 10^-2m, q = 1.6 x 10^-19C, m = 9.1 x 10^-31 kg and $\varepsilon$₀ = 8.85 x 10^-12C²/Nm²

Therefore,

v² = [2 (1.6 x 10^-19C) (1.37 x 10^-2m)(1.61 x 10^-7C/m²) ]/(9.1 x 10^-31 kg)(8.85 x 10^-12C²/Nm²)

= (7.05 x 10^-28 C²/m)/(80.53 x 10^-43 kg C² m³ s^-2) = 87.54 x 10¹² (m/s)²

Hence, v = 9.35 x 10⁶ m/s.

(2)