Tutorial 4+ans

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BWC21103 ELECTROMAGNETISM
Tutorial 4
1. Let us illustrate the use the vector form of Coulomb’s law by locating a charge of Q1 = 3 x
10-4 (C) at (1, 2, 3) at a charge Q2 = -10-4 (C) at (2, 0, 5) in a vacuum. We desire the force
exerted on Q2 by Q1. (Ans: 𝐹12 = −10π‘Žπ‘₯ + 20π‘Žπ‘¦ − 20π‘Žπ‘§ ).
Solution
2. Point charges 1 mC and -2mC are located at (3, 2, -1) and (-1, -1, 4), respectively. Calculate
the electric force on a 10 nC charge located at (0, 3, 1) and the electric field intensity at that
point. (Ans: 𝐸 = −650.7π‘Žπ‘₯ − 381.7π‘Žπ‘¦ + 750,6π‘Žπ‘§ kV/m)
Solution
3. A circular ring of radius a carrier a uniform charge ρL C/m and is placed on the xy-plane
with axis the same as the z-axis.
(a) Show that
𝜌𝐿 π›Όβ„Ž
𝐸(0, 0, β„Ž) =
π‘Ž
Μ‚
2πœ€0 [β„Ž2 + 𝛼 2 ]3⁄2 𝑧
𝛼
(b) What values of h gives the maximum value of E? (Ans: β„Ž = ± )
√2
𝑄
(c) If the total charge on the ring is Q, find E as α→ 0. (Ans:𝐸 = 4πœ‹πœ€
Solution
0π‘Ÿ
2
π‘Ž
Μ‚)
𝑧
4. The finite sheet 0 < x < 1, 0 < y < 1 on the z =0 plane has a charge density πœŒπ‘  =
3
π‘₯𝑦(π‘₯ 2 + 𝑦 2 + 25)2 nC/m2. Find
(a) The total charge on the sheet. (Ans: Q = 33.15 nC)
(b) The electric field at (0, 0, 5). (Ans: (-1.5, -1.5, 11.25) V/m)
(c) The force experienced by a – 1 mC charge located at (0, 0, 5) (Ans: (-1.5, -1.5, 11.25)
mN)
Solution
5. Determine D at (4, 0, 3) if there is a point charge -5π mC at (4, 0, 0) and a line charge 3π mC
along the y axis. (Ans: D = DQ + DL, DQ = -0.138az mC/m2, DL = 0.24ax + 0.18 az mC/m2, D
= 240ax+42az µC/m2)
Solution
6. A non-conducting ring of radius R with a uniform charge density, D and a total charge Q is
lying in the xy-plane, as shown in figure below. Compute the electric field at a point P,
located at a distance z from the center of the ring along its axis of symmetry. (Ans: 𝐸𝑧 =
1
4πœ‹πœ–0
𝑄𝑧
2
(π‘Ÿ +𝑧 2 )3⁄2
Electric field at P due to the charge element dq
Solution
7. In the classical model of the hydrogen atom, the electron revolves around the proton with a
radius of r = 0.53 x 10-10 m. The magnitude of the charge of the electron and proton is e = 1.6
x 10-19 C.
(a) What is the magnitude of the electric of the electric force between the proton and the
electron? (Ans: 8.2 x 10-8 N)
(b) What is the magnitude of the electric field due to the proton at r? (Ans: 5.76 x 1011 N/C)
(c) What is ratio of the magnitude of the electrical and gravitational force between electron
and proton? Does the result depend on the distance between the proton and the electron?
(Ans: 2.2 x 1039 )
(d) In the light of your calculation in (b), explain why electrical forces do not influence the
motion of planets.
Solution
8. A non-conducting rod of length, l with a uniform charge density, D and a total charge Q is
lying along the x-axis, as illustrated in figure below. Compute the electric field at a point P,
𝜌
(sin πœƒ2 − sin πœƒ1 )
located at a distance, y off the axis of the rod. (Ans: 𝐸𝑦 =
4πœ‹πœ€0 𝑦
Solution
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