The electric field intensity at a point 2m from the surface of a uniformly charged conducting sphere is 1920 N/C. At a point 4m from its center, the electric field intensity is 3000 N/C.
a.) Determine the radius of the sphere. Determine the electric field intensity at a point:
(b.) 2m from its center,
(c.) 6m from its center, and
(d.) 4m from its surface.
3m
0
1333.33
979.60
To solve this problem, we can use two key concepts:
1. Gauss's Law: Gauss's Law states that the electric field passing through a closed surface is proportional to the electric charge enclosed by that surface divided by the permittivity of the medium. For a uniformly charged conducting sphere, the electric field at any point outside the sphere is the same as if all the charge were concentrated at its center.
2. Coulomb's Law: Coulomb's Law describes the force between two electric charges. It states that the force between two charges is directly proportional to their magnitudes and inversely proportional to the square of the distance between them.
Now let's solve this problem step by step:
a.) To determine the radius of the sphere, we need to use the information given about the electric field intensities at two different distances. We can calculate the charge Q enclosed by a Gaussian surface placed at each of the given distances using Gauss's Law.
At 2m from the surface:
Electric field intensity (E) = 1920 N/C
Distance from the center (r) = 2m
Using Gauss's Law, we can write:
EΔA = (Qenclosed)/ε0
Substituting the given values:
1920 N/C * 4π(2m)^2 = Qenclosed / (8.85 x 10^-12 C^2/N.m^2)
Simplifying, we get:
Qenclosed = (1920 N/C * 4π * 4m^2 * 8.85 x 10^-12 C^2/N.m^2)
Now, at 4m from the center:
Electric field intensity (E) = 3000 N/C
Distance from the center (r) = 4m
Again, using Gauss's Law:
3000 N/C * 4π(4m)^2 = Qenclosed / (8.85 x 10^-12 C^2/N.m^2)
Simplifying, we get:
Qenclosed = (3000 N/C * 4π * 16m^2 * 8.85 x 10^-12 C^2/N.m^2)
Since the charge enclosed should be the same for both cases, we can equate the two expressions for Qenclosed and solve for the radius (r) of the sphere.
(1920 N/C * 4π * 4m^2 * 8.85 x 10^-12 C^2/N.m^2) = (3000 N/C * 4π * 16m^2 * 8.85 x 10^-12 C^2/N.m^2)
Simplifying and solving for r, we find:
r^2 = (3000 N/C * 16m^2) / (1920 N/C)
r^2 = 25
r = 5m
Therefore, the radius of the sphere is 5m.
b.) To determine the electric field intensity at a point 2m from the center of the sphere, we can apply Coulomb's Law.
Electric field intensity (E) is directly proportional to the charge (q) and inversely proportional to the square of the distance (r) between the charges:
E = k * (q / r^2)
Given that the charge is uniformly distributed over the sphere, we can assume that the charge enclosed by the Gaussian surface at a distance of 2m from the center is the same as the charge enclosed by a sphere with a radius of 5m.
Using Coulomb's Law, we can write:
E = k * (Qenclosed / r^2)
E = k * (Q / 2^2)
E = k * (Q / 4)
Substituting the values:
E = (9 x 10^9 N.m^2/C^2) * (Q / 4)
We already calculated the charge enclosed, Qenclosed, which is the same as Q, as (1920 N/C * 4π * 4m^2 * 8.85 x 10^-12 C^2/N.m^2).
Plugging in the calculated value of Qenclosed into the equation, we can solve for E.
E = (9 x 10^9 N.m^2/C^2) * [(1920 N/C * 4π * 4m^2 * 8.85 x 10^-12 C^2/N.m^2) / 4]
Simplifying the expression gives us the electric field intensity at a point 2m from the center of the sphere.
c.) To determine the electric field intensity at a point 6m from the center of the sphere, we can use the same approach as in part (b).
Using Coulomb's Law, we have:
E = k * (q / r^2)
Since the charge enclosed is the same as before, we can substitute the previously calculated value of Qenclosed into the equation:
E = (9 x 10^9 N.m^2/C^2) * [(1920 N/C * 4π * 4m^2 * 8.85 x 10^-12 C^2/N.m^2) / (6m)^2]
Simplifying the expression gives us the electric field intensity at a point 6m from the center of the sphere.
d.) To determine the electric field intensity at a point 4m from the surface of the sphere, we can use the electric field intensity formula for a point outside the sphere. For a uniformly charged conducting sphere, the electric field at any point outside the sphere is the same as if all the charge were concentrated at its center.
Therefore, the electric field intensity at a point 4m from the surface is the same as the electric field intensity at a point 6m from the center, which we calculated in part (c).