A hollow conducting sphere with inner radius R and outer radius 2R has a non-uniform volume charge distribution in the region R<r<2R given by ρ(r)=(4C/m5)r2. If R=0.18m, what is the total charge (in C ) of the sphere?
To find the total charge of the sphere, we need to integrate the charge density ρ(r) over the volume of the sphere.
The charge density ρ(r) is given by ρ(r) = (4C/m^5)r^2, where r is the distance from the center of the sphere.
To integrate the charge density over the volume of the sphere, we can express the volume element dV in terms of the radial coordinate r. The volume of a hollow sphere between radii R and 2R is given by:
dV = 4πr^2 dr (from the volume element of a spherical shell)
We can substitute this expression for dV into the charge integral:
Q = ∫ρ(r) dV
= ∫(4C/m^5)r^2 * 4πr^2 dr
= 16πC/m^5 ∫r^4 dr
Integrating with respect to r, we obtain:
Q = 16πC/m^5 * [1/5 * r^5] | from R to 2R
= 16πC/m^5 * (1/5 * (2R)^5 - 1/5 * R^5)
= 16πC/m^5 * (1/5 * 32R^5 - 1/5 * R^5)
= 16πC/m^5 * (32/5 R^5 - 1/5 R^5)
= 16πC/m^5 * (31/5 R^5)
Now we substitute R = 0.18m into the equation:
Q = 16πC/m^5 * (31/5 * (0.18m)^5)
= 16πC/m^5 * (31/5 * 0.00004m^5)
= 16πC/m^5 * 0.000019377m^5
≈ 0.001198C
Therefore, the total charge of the sphere is approximately 0.001198 Coulombs (C).