Question

Two parallel plates having charges of equal magnitude but opposite sign are separated by 21.0 cm. Each plate has a surface charge density of 39.0 nC/m2. A proton is released from rest at the positive plate. (a) Determine the magnitude of the electric field between the plates from the charge density.

Answer 1

Answer:

E = 3.45*10^-19 N/C

Explanation:

a) The electric field between two parallel plates id given by the following formula:

$E=\frac{\sigma}{\epsilon_o}$           (1)

where:

σ: surface charge density of the plates = 39.0nC/m^2

εo: dielectric permittivity of vacuum = 8.85*10^-12 C/Nm^2

You replace these values in the equation (1):

$E=\frac{39.0*10^{-9}C/m^2}{8.85*10^{-12}C^2/Nm^2}\\\\E=3.45*10^{-19}\frac{N}{C}$

The electric field in between the parallel plates is 3.45*10^-19 N/C