In a spherical region, the voltage is measured to be spherically symmetrical, with V = VCr) = wrg • The constants, wand g, are positive.
a.
Find the radial electric field.
b.
Use Gauss' Law to find the charge enclosed in a sphere of radius r.
c.
Find the charge enclosed by a sphere ofradius r+dr.
d.
Find the differential charge enclosed in the annular region between two
concentric spheres of radii rand r+dr.
e.
Find the differential volume ofthe annular region between two concentric
spheres of radii rand r+dr.
f.
Find the charge density, p = per) =?
E = -dV/dr = - w g which is constant
so at any r find the charge inside from Gauss
E * (4/3)pi r^2 = Q/eo
Q = (4/3) pi r^2 eo E
so
Q(r) = -w g (4/3) pi eo r^2
Q(r+dr) = - w g (4/3) pi eo (r^2+2 r dr + dr^2)
so charge between spheres =
Q(r+dr)-Q(r) = -w g (4/3) pi eo (2 r dr + dr^2)
as dr^2 is << r dr as dr -->0
Q(r+dr) - Q(r) =- w g (4/3) pi eo (2 r dr)
volume between = area of sphere * dr = (4/3) pi r^2 dr
so charge density between = - 2 w g eo /r
thank you very much Damon...you help me alot. thanks
a. To find the radial electric field, we need to find the gradient of the potential function V(r). The electric field E(r) is the negative gradient of V(r) with respect to r.
E(r) = -grad(V(r))
In this case, since V = wrg, the potential function is only dependent on the radial distance r. Therefore, we can simply differentiate V with respect to r to find the electric field.
E(r) = dV/dr = d(wrg)/dr
E(r) = wg
So, the radial electric field in this spherical region is E(r) = wg.
b. Using Gauss' Law, we can find the charge enclosed in a sphere of radius r by integrating the electric field over a closed surface. Gauss' Law states that the electric flux through a closed surface S is equal to the charge enclosed divided by the electric constant (ε0).
∮ E · dA = qenc / ε0
For a spherical Gaussian surface of radius r, the electric field E is constant and perpendicular to the surface. Thus, the flux equation simplifies to:
E * 4πr² = qenc / ε0
Solving for the charge enclosed qenc:
qenc = E * 4πr² * ε0 = wg * 4πr² * ε0
c. To find the charge enclosed by a sphere of radius r+dr, we consider the differential charge within the differential volume enclosed by the two spheres. The differential charge is given by the charge density multiplied by the differential volume.
dQ = p(r+dr) * dV
Substituting the differential volume expression dV = 4πr²dr:
dQ = p(r+dr) * 4πr²dr
d. To find the differential charge enclosed in the annular region between two concentric spheres of radii r and r+dr, we subtract the differential charge inside the smaller sphere from the differential charge inside the larger sphere.
dQannular = dQ(r+dr) - dQ(r)
By substituting the expression for dQ from part c, we can calculate the differential charge in the annular region.
dQannular = [p(r+dr) * 4π(r+dr)²dr] - [p(r) * 4πr²dr]
e. The differential volume of the annular region between two concentric spheres of radii r and r+dr is given by the difference in the volumes of the two spheres.
dVannular = V(r+dr) - V(r)
The formula for the volume of a sphere is V = (4/3)πr³. Substituting this expression, we find:
dVannular = [(4/3)π(r+dr)³] - [(4/3)πr³]
f. The charge density p(r) is defined as the charge per unit volume. To find the charge density, we need to divide the charge enclosed by the volume.
p(r) = qenc / V(r)
By substituting the expression for qenc from part b and V(r) as the volume of a sphere:
p(r) = (wg * 4πr² * ε0) / [(4/3)πr³]
p(r) = 3wgε0 / r