ELEC210: Problem set 1
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ELEC210: Problem set 1 (Lectures 1, 2, 3)
1. Find the vector between the two points A = (2, -4, 1) and B = (0, -2, 0).
2. If A is the vector from the origin (0, 0, 0) to the point (3, -4, 0), and B is the vector from the origin (0, 0, 0) to the point (-1, 3, 4), find A ∙ B.
3. Given vectors A = ෝ(x) + ෝ(y), B = ෝ(x) + 2ො(z) and C = 2ෝ(y) + ො(z), find A × B × C.
4. Consider the rectangular object shown on Page 14 of Lecture 2. Calculate the mass of the object if the mass density has distribution p(x, y, z) = 5 xଶ y kg / m3
5. The electric field E is 100 V m-1 at a distance of 2 m from a point charge Q.
a) Find Qif the surrounding material has relative dielectric constant εr =1.5.
b) Is E radially inward or outward? Give reasons for your answer.
6. A charge 30 nC is located at the point (x, y, z) = (0, 2, 0) m and a charge of –30 nC is located at the point (0, -2, 0) m. Assume that the surrounding material has relative dielectric constant εr =1.
a) Find E at point A = (4, 2, 4) m
b) Find E at point B = (4, 0, 4) m
c) Find E at point C = (4, -2, 4) m
7. Find the electric field E for the two following cases
a) When the scalar electric potential is given by V = 2xଶy + 4z
b) When the scalar electric potential is given by V = 3xeିଶ௫
8. A long straight wire in air carries a charge of 125 nC m-1 . Find E at a distance of 3 m from the wire.
9. A uniform line charge, infinite in extent, has charge density of p
= 45 nC mିଵ . If the line charge is oriented along the z axis, find E at the point (6, 8, 3) m.
10. Calculate the distance at which the absolute potential of a 120 nC point charge equal to 1 kV for the two cases (a) and (b)
a) in a vacuum
b) in a medium with εr = 4.
11. Two point Qଵ= 50 mC and Qଶ= 10 mC are located at the points (-1, 1, -3) and (3, 1, 0) respectively. Find the force on Qଵ .
12. Two long parallel conductors of a d.c transmission line separated by 2 m have charges of 5 μC m -1 of the same sign. Both lines are more than 10 m above the ground.
Calculate the magnitude of the electric field 4 m directly below one of the wires.
(This problem is nearly identical to Example 2.3 in Lecture 2, with the difference that the two
lines both have charge densities with the same sign. You should follow the same logic as shown in the lectures to find the answer.)
Numerical answers to selected problems: full solutions available later in the semester
5. 66.73 nC, radially outward.
6. (a) E = 2.71x(ˆ) - 3.25y(ˆ) + 2.71z(ˆ) (V m-1); (b) -5y(ˆ) (V m-1); (c) = -2.71x(ˆ) - 3.25y(ˆ) - 2.71z(ˆ) (V m-1)
7. (a) E = -4xyx(ˆ) - 2y(ˆ) - 4z(ˆ) (V m-1); (b) E = 6e-2x x(ˆ) (V m-1)
8. 749.3 V m-1
10. (a) 1.08 m; (b) 0.27 m
2023-08-28