Question

010. A proton is released from rest at the positive plate of a parallel plate capacitor. The charge per unit area on each plate is 1.8x10 Cl, and the plates are separated by a distance of 1.5x10 m. (a) (5 points) What is the magnitude of the electric field between the two plates? (b) (5 points) What is the potential difference between the two plates?

Answer 1

Charge per unit area of plate is $\sigma =1.8*10^{-7}\,C/m^2$

Distance between the plates $d=1.5*10^{-2}\,m$

(a)

Magnitude of electric field between the plates $E=\frac{\sigma }{\epsilon_0}=\frac{1.8*10^{-7}}{8.85*10^{-12}}=20339\,N/C=2.034*10^4\,N/C$

(b)

Potential difference between the plates

$V=Ed=20339*1.5*10^{-2}=305.1\,V$