To find the voltage between two equipotential points, we can use the formula:
V = Ed
Where:
V is the voltage between two points
E is the electric field intensity
d is the distance between the two points
First, let's find the electric field intensity using the electric flux density:
E = D / ε0
Where:
E is the electric field intensity
D is the electric flux density
ε0 is the permittivity of free space
Given:
D = 1.6 nC/m^2
ε0 = 8.854 x 10^-12 C^2 / N * m^2
Plugging in the values:
E = (1.6 x 10^-9 C/m^2) / (8.854 x 10^-12 C^2 / N * m^2)
E ≈ 180.92 N/C
Next, let's find the distance between the two equipotential points:
d = 2.54 cm
d ≈ 0.0254 m
Now, we can calculate the voltage:
V = E * d
V ≈ (180.92 N/C) * (0.0254 m)
V ≈ 4.6076 V
Rounded to three decimal places, the voltage between the two equipotential points is approximately 4.607 V.
None of the given options match the calculated voltage exactly.
A certain homogeneous slab of lossless dielectric material is characterized by an electric susceptibility of 0.12 and carries a uniform electric flux density within it of
1.6nC/m
2
. Find the voltage between two equipotential
2.54 cm
apart. a.
4.656 V
b.
4.216 V
c.
4.919 V
d.
4.097 V
1 answer