Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0 cm. Each plate has a surface charge density of 49.0 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.
b) Determine the potential difference between the plates.
Answer:
a
The Electric Field between the two plate is [tex]E = 5.536*10^3 N/C[/tex]
b
The potential difference between the plate is V = [tex]609 \ Volts[/tex]
Explanation:
From the we are given that
The separation between the plate is [tex]d = 11.0 cm[/tex]
The surface charge density is [tex]\sigma = 49.0 n C/m^2 = 49.0*10^{-9}[/tex]
Generally Electric field between the plate is mathematically given as
[tex]E = \frac{\sigma }{\epsilon_0}[/tex]
Note that [tex]\epsilon_0[/tex] is the permitivity of free space and its value is [tex]8.85 *10 ^{-12} C^2 /N \cdot m^2[/tex]
Now substituting values we have
[tex]E = \frac{49*10^{-9}}{8.85*10^{-12}}[/tex]
[tex]5.537*10^3 N/C[/tex]
Generally Potential difference between the plate is mathematically given as
[tex]V =Ed[/tex]
Where E is the electric field which is [tex]E = 5.536*10^3 N/C[/tex]
Substituting value we have
V = [tex]= (5.537*10^3 N/C) (11.0*10^{-2}) =609 V[/tex]
