1. Lecture 7 Practice Problems
Prof. Viviana Vladutescu

2. Divergence Theorem and Gauss’s Law
Suppose D = 6rcos φ aφ C/m2.(a) Determine the charge density at the point (3m, 90, -2m). Find the total flux through the surface of a quartered-cylinder defined by 0 ≤ r ≤ 4m, 0 ≤ φ ≤ 90, and -4m ≤ z ≤ 0 by evaluating (b) the left side of the divergence theorem and (c) the right side of the divergence theorem.
(a)
(b)

3. note that the top, bottom and outside integrals yield zero since
there is no component of D in the these dS directions.
So,
(c)

4. Electric Potential The potential field in a material with εr= 10.2
is V = 12 xy2 (V). Find E, P and D.

5. Boundary Conditions
For z ≤ 0, er1 = 9.0 and for z > 0, er2 = 4.0. If E1 makes a 30 angle with a normal to the surface, what angle does E2 make with a normal to the surface?

6. Therefore
also
and after routine math we find
Using this formula we obtain for this problem q2 = 14°.

7. 1. A spherical capacitor consists of a inner conducting sphere of radius R1=3cm, and surface charge density ρs=2nC/m2 and an outer conductor with a spherical inner wall of radius R2=5cm . The space in between is filled with polyethylene.
a) Determine the electric field intensity from ρs between plates(15 points)
b) Determine the voltage V12 (10 points)
c) Determine the capacitance C (10 points)
2. Consider a circular disk in the x-y plane of radius 5.0 cm. Suppose the charge density is a function of radius such that ρs = 12r nC/cm2 (when r is in cm). Find the electric field intensity a point 20.0 cm above the origin on the z-axis.(25 points)

8. 3. Assume the z=0 plane separates two lossless dielectric regions, where the first one is air and the second one is glass. If we know that the electric field intensity in region 1 is E1 =2y ax -3x ay+(5+z) az,
a)Find E2 and D2 at the boundary side of region 2? (20 points)
b)Can we determine E2 and D2 at any point in region 2? Explain.(5 points)
4. Two spherical conductive shells of radius a and b (b > a) are separated by a material with conductivity σ. Find an expression for the resistance between the two spheres.(20 points)

9. Problem 2 For a ring of charge of radius a, Now we have rL=rsdr and
where rs = Ar nC/cm2.
Now the total field is given by the integral:

10. This can be solved using integration by parts, where u = r,
du = dr,
This leads to
Plugging in the appropriate values we arrive at E = 6.7 kV/cm az.

11. Problem 3
At the z=0 plane
From the boundary condition of two dielectrics we get

12. z=0

13. Problem 4- Hint J σ
Vab a b I l

14. Resistance of a straight piece of homogeneous material of a uniform cross section for steady current

15. Resistances connected in series
Resistances connected in parallel

16. Problem 4 First find E for a < r < b, assuming +Q at r = a and
–Q at r = b. From Gauss’s law:
Now find Vab:

17. Now can find I:
Finally,