Determining how the charges distribute on the surface of a conductor is, in general, a very difficult problem. We know that if we charge a conductor the charges go to the surface and redistribute so that the electric field in the conductor vanishes. One of the few shapes for which this distribution can be determined analytically is the ellipsoid.
x^2/a^2 + y^2/b^2 + z^2/c^2=1.
Here, a, b and c are the ellipsoid's semi-axes. One can prove that for an ellipsoidal conductor the surface charge density is given by
σ(x,y,z)=Q/(4 pi abc) X 1/root(x^2/a^4 + y^2/b^4 + z^2/c^4)
where Q is the the net charge of the conductor. Note that if we set a=b=c=R we obtain the uniform charge distribution Q/(4 pi R^2), corresponding to a spherical conductor. Suppose that we measure the electric field near the surface of a charged ellipsoid with Q=1nC, a=2cm,b=5cmand c=3cm. What is the maximum value in in volts per meter of the electric field?
Details and assumptions
k=1/(4 pi ϵ0) =9×10^9m/F
To find the maximum value of the electric field near the surface of a charged ellipsoid, we need to use the formula for electric field intensity (E) near a charged conductor:
E = σ / ϵ0
Here, σ is the surface charge density and ϵ0 is the permittivity of free space.
In the question, it is mentioned that the surface charge density for an ellipsoidal conductor is given by:
σ(x,y,z) = Q / (4πabc) × (1 / √(x^2/a^4 + y^2/b^4 + z^2/c^4))
Given:
Q = 1 nC (net charge of the conductor)
a = 2 cm
b = 5 cm
c = 3 cm
k = 9×10^9 Nm^2/C^2 (electrostatic constant)
To find the maximum electric field, we need to determine the maximum value of σ(x,y,z) and plug it into the electric field formula.
Since we have an ellipsoid, we need to find the maximum value of σ(x,y,z) on the surface of the ellipsoid. To do that, we substitute the values of x, y, and z with the semi-axes a, b, and c in the equation for σ(x,y,z).
σ_max = Q / (4πabc) × (1 / √(a^2/a^4 + b^2/b^4 + c^2/c^4))
Simplifying further:
σ_max = Q / (4πabc) × (1 / √(1/a^2 + 1/b^2 + 1/c^2))
Now, we can substitute the given values to find σ_max:
σ_max = (1 nC) / (4π × 2 cm × 5 cm × 3 cm) × (1 / √(1/(2 cm)^2 + 1/(5 cm)^2 + 1/(3 cm)^2))
Converting the centimeters to meters:
σ_max = (1 nC) / (4π × 0.02 m × 0.05 m × 0.03 m) × (1 / √(1/(0.02 m)^2 + 1/(0.05 m)^2 + 1/(0.03 m)^2))
Now we can calculate σ_max using a calculator.
Once we have σ_max, we can calculate the maximum electric field (E_max) using the formula:
E_max = σ_max / ϵ0
E_max = (σ_max) / (9×10^9 Nm^2/C^2)
Plugging in the calculated value of σ_max and the value of k, we can find E_max in volts per meter.
Remember to always double-check your calculations and units to ensure accuracy.