Learning Objectives:

• Learn about different coordinate systems, such as Cartesian coordinates and polar coordinates, and how they are used to represent spatial positions.

Types of Co-ordinate Systems:

1. Rectangular or Cartesian Coordinate system: (x,y,z)
2. Cylindrical coordiante system: (r, $\theta$, z)
3. Spherical coordinate system (r, $\theta$, $\phi$)

Transformation:

• Cartesian into cylindrical:
\begin{align*} r &= \sqrt{x^2 + y^2}\\ \theta &= \tan^{-1}\left(\dfrac{y}{x}\right)\\ z &= z \end{align*}
• Cartesian into cylindrical:
\begin{align*} r &= \sqrt{x^2 + y^2 + z^2}\\ \theta &= \tan^{-1}\left(\dfrac{y}{x}\right)\\ \phi &= \cos^{-1}\left(\dfrac{z}{r}\right) \end{align*}

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Solved Example: 9994-01

The electric field strength at distant point, P, due to a point charge, +q, located at the origin, is 100 $\mu$ V/m. If the point charge is now enclosed by a perfectly conducting metal sheet sphere whose center is at the origin, then the electric field strength at the point, P, outside the sphere, becomes:

A. Zero

B. 100 $\mu$ V/m

C. – 100 $\mu$ V/m

D. 50 $\mu$ V/m

The point charge +q will induce a charge – q on the surface of metal sheet sphere. Using Gauss’s law, the net electric fulx passing through a closed surface is equal to the charge enclosed = + q – q = 0 D = 0, E = 0 at point P.

Correct Answer: A

Solved Example: 9994-02

A metal sphere with 1 m radius and surface charge density of 10 Coulombs / m$^2$ is enclosed in a cube of 10 m side. The total outward electric displacement normal to the surface of the cube is:

A. 40 $\pi$ Coulombs

B. 10 $\pi$ Coulombs

C. 5 $\pi$ Coulombs

D. None of the above

The sphere is enclosed in a cube of side = 10m. using Gauss’s law, the net electric flux flowing out through a closed surface is equal to charge enclosed. \[\begin{aligned} \iint_s \bar(D) \cdot \bar{da} &= Q (enclosed)\\ &= P_S 4 \pi r^2\\ &= 10 \times 4 \pi\\ &= 40 \pi Coulombs\end{aligned}\]

Correct Answer: A

Solved Example: 9994-03

The thumb in Fleming's left hand rule indicate:

A. Motion

B. Current

C. Field

D. None from the above

Correct Answer: A

Solved Example: 9994-04

Electric flux is a _______ field, and its density is a _______ field.

A. Vector, vector

B. Scalar, vector

C. Vector, scalar

D. Scalar, scalar

Correct Answer: B

Solved Example: 9994-05

Find H = ___________ A/m at the center of a circular coil of diameter 1 m and carrying a current of 2 A.

A. 0.6366

B. 0.1636

C. 6.366

D. 2

Correct Answer: D

Solved Example: 9994-06

The force between two charges is 200 N. If the distance between the charges is doubled, the force will be _______.

A. 400 N

B. 100 N

C. 50 N

D. 200 N

Correct Answer: C

Solved Example: 9149-01

The divergence theorem converts:

A. Line to surface integral

B. Surface to volume integral

C. Volume to line integral

D. Surface to line integral

The divergence theorem for a function F is given by \[ \iint F. dS = \iiint Div (F). dV\] Thus it converts surface(double integral) to volume (triple integral).

Correct Answer: B

Solved Example: 9149-02

Evaluate Gauss law for D = 5r$^2$/4 in spherical coordinates with r = 4m and $\theta$ = $\dfrac{\pi}{4}$ as surface integral.

A. 600

B. 588.9

C. 577.8

D. 599.7

\[\iint D. ds =\iint \dfrac{5r^2}{4}. ds\] In spherical co-ordinates, $ds$= surface area = $r^2 \sin \theta d\theta d\phi$ \begin{align*} &= \iint \dfrac{5r^2}{4}. r^2\ \sin \theta d\theta\ d\phi\\ &= \iint \dfrac{5r^4}{4}\ \sin \theta d\theta\ d\phi \end{align*} Given r = 4 and it remains constant throughout the surface on a sphere, \begin{align*} &= \dfrac{5r^4}{4} \times \int_0^{\dfrac{\pi}{4}} \sin \theta d\theta \int_0^{2\pi} d\phi\\ &= \dfrac{5(4)^4}{4} \times \int_0^{\dfrac{\pi}{4}} \sin \theta d\theta \int_0^{2\pi} d\phi\\ &= 320 \times \left(-\cos \theta\right)_0^{\frac{\pi}{4}} [\phi]_0^{2\pi}\\ &= (-320) \times \left(\cos \dfrac{\pi}{4} - \cos 0\right)[2\pi]\\ &= (-320) \times \left(-0.29289\right)[2\pi]\\ &= 588.9 \end{align*}

Correct Answer: B