Question

Part A

If a capacitor has opposite 5.0 μC charges on the plates, and an electric field of 2.4 kV/mm is desired between the plates, what must each plate's area be?

Express your answer using two significant figures and include the appropriate units.

Part B

How much kinetic energy will an electron gain (in joules) if it accelerates through a potential difference of 1.69×10⁴ V ?

Part C

How much kinetic energy will an electron gain (in eV) if it accelerates through a potential difference of 1.69×10⁴ V ?

Answer 1

1. Charge on capacitor = q = 5.0 μC

=5.0 x 10^-6 C

E = electric field = 2.4 kV/mm

= 2.4 *1000 / 10^-3

= 2.4 x10⁶ V/m

q = C*V = ( ϵo*A /d)*V =( ϵo*A*E)

Area , A = q /( ϵo*E)

=5.0 x 10^-6 C / (8.86 x 10^-12* 2.4 x10⁶ )

= 1.0632 x 10^-10 m²

2. kinetic energy gained by electron

=work done

= potential difference * charge of electron

=1.69×10⁴ V * 1.6 x 10^-19

= 2.704 x 10^-15 joules

3. kinetic energy gained by electron

=work done

= potential difference * charge of electron

   =1.69×10⁴ V * 1e=1.69×10⁴ eV