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Unit - 1

Cartesian

Q1) Obtain the unit vector in the direction from the origin towards the point P (3, -3, -2)

A1)

The origin O (0,0,0) while P (3, -3, -2) hence the distance vector is

= (3-0) + (-3-0) +(-2-0) = 3 -3 -2

= (3) 2 + (-3) 2 + (-2) 2 = 4.6904

Hence the unit vector along the direction OP is

OP= / | | = 3 -3 -2 / 4.6904 =

= 0.6396 -0.63963 -0.4264

Q2) Given the three points in cartesian co-ordinate system as A (3, -2,1) B (-3, -3,5), C (2,6, -4)

Find

i) The vector from A to C

ii) The unit vector from B to A

A2) The position vectors for the given points are:

= 3 - 2+.

= -3 - 3+5 .

= 2+6-4.

i) The vector from A to C is

= - = [ 2-3] + [6– (-2)]+ [-4-1] .

= - + 8-5 .

ii) For unit vector from B to A obtain the distance vector first

= -

= [ 3 – (-3)] + [ (-2) – (-3)] + [ -4-1]

= 6 + -4 .

| | = (6) 2 + (1) 2 + (-4)2 = 7.2801.

BA = / || = 6 + -4 ./ 7.2801

Q3) Given the three points in cartesian co-ordinate system as A (3, -2, 1) B (-3, -3,5), C (2,6, -4)

Find

i) The distance from B to C

ii) The vector from A to the mid-point of the straight-line joining B to C.

A3) The position vectors for the given points are:

= 3 - 2+.

= -3 - 3+5 .

= 2+6-4.

i) For distance between B and C

= - = [ 2 – (-3] + [6-(-3)] + [-4) – (-5)] = 5 + 9 – 9

Distance BC = 2 + (9) 2 + (-9) 2 = 13.6747

ii) Let B (x1, y1, z1) and C (x2, y2, z2) then the co-ordinates of the mid-point BC are (x1+x2/2, y1+y2/2, z1+z2/2)

Q4) A charge Q1=-20µC is located at P (-6,4,6) and a charge Q2=50µC is located at R (5,8, -2) in a free space. Find the force exerted on Q2 by Q1 in vector form. The distances given are in meters.

A4) From the co-ordinates of P and R the respective position vectors are:

= -6 + 4 + 6

and

= 5 + 8 -2

The force on Q2 is given by

= Q1 Q2 / 4 π R212.

= = - = [ 5 – (-6)] + (8-4) + [-2-(-6) ]

= 11 + 4 - 8

|R12| = (11) 2 + (4) 2 +(-8) 2 = 14.1774

= / || = 11 + 4 -8 / 14.1774

= -20 x 10 -6 x 50 x 10 -6 / 4 π x 8.854 x 10 -12 x (14.1774) 2 [

= -0.0447 [ 0.7758 + 0.2821 - 0.5642 ]

= -0.0346 - 0.01261 + 0.02522 N

This is the required force exerted on Q2 by Q1

The magnitude of the force is

| = (0.0346) 2 + (0.01261) 2 + (-0.02522) 2 = 44.634mN

Q5) A scalar field is given by

Find its gradient at P (0,1,1) for Cartesian, P ( for cylindrical and P for the spherical.

A5)

the +9.0689

At the

Q6) Find the expression of energy stored in magnetic field?

A6) Let us consider the magnetic field is due to electromagnet that is a coil of several no. turns.

This coil or inductor is carrying current I when it is connected across a battery or voltage source through a switch.

Figure. Inductor circuit

Suppose battery voltage is V volts, value of inductor is L Henry, and current I will flow at steady state.

When the switch is ON, a current will flow from zero to its steady value. But due to self- induction an induced voltage appears which is

E = -L dI/dt ---------------------------------------------------------------------(1)

Here E always in the opposite direction of the rate of change of current.

The energy or work done due to this current passing through this inductor is U. As, the current starts from its zero value and flows against the induced emf E, the energy will grow up gradually from zero value to U.

dU = W. dt, where W is the small power and W = – E.I

So, the energy stored in the inductor is given by

dU = W. dt = -E. I dt = L dI/dt. I dt = LIdI

Now integrate the energy from 0 to its final value.

U = = dI = ½ LI 2

Again

L = o N 2 A /l

where N is the number of turns of the coil,

A is the effective cross-sectional area of the coil and l is the effective length of the coil.
Again,

I = H.l/N

Where, H is the magnetizing force, N is the number of turns of the coil and l is the effective length of the coil.

I = B.l/

Now putting expression of L and I in equation of U, we get new expression i.e.

U = μoN2 A /l. B.L/ μo N / 2 = B 2 Al/ 2 μo

Q7) What are scalar and vector fields?

A7) If at every point of a region, there is corresponding value of some physical function, the region is called a field. Mathematically, a field is a function which describes a physical quantity at all points in space. Field is classified as scalar or vector.

i) Scalar field: If the value of the physical function at every point is scalar quantity, then the field is scalar field for example temperature of atmosphere density of the nonhomogeneous body and electric potential scalar field.

ii) Vector field: If the value of the physical function at every point is vector quantity, the field is vector field for example the wind velocity of atmosphere, gravitational force and electric field intensity are vector fields.

Q8) Write the expressions for line, surface and volume integral?

A8) Line Integrals

It is defined as any integral which is to be determined along a curve. Line integral can be defined in terms of limits of sum as are the integrals of ordinary calculus.

Above mentioned are some examples of line integrals calculated from point a to point b. In first case each element of length dl on the curve is multiplied by the local value A scalarly and then these products are summed to get the value of the integral.

Line integral in terms of cartesian coordinate can be represented as

Surface Integrals

Fig: A two-sided surface

Differential Distances:

Differential Surfaces:

Different Volume:

Volume Integrals

If a closed surface in space enclosing a volume V is considered then the integrals

Are the examples of volume integrals. Since volume is a three-dimensional quantity and hence three integrals.

Q9) Draw a block diagram to show the relation between line surface and volume integral. How they are related by Green’s theorem?

A9)

Q10) Define Coulombs Law?

A10) It states that force between two-point charges Q1 and Q2

Acts along the line joining the two-point charges.

Is directly proportional to the product of Q1Q2 of the two charges

Is inversely proportional to the square of the distance between them.

Figure. Two point charge

Consider two point charges Q1 and Q2 as shown in the figure seperated by distance R. The charge Q1 exerts a force on Q2 while Q2 also exerts a force on Q1. The force acts along the line joining Q1 and Q2. The force exerted between them is repulsive if the chrages are of same polarity while it is attractive if the charges are of different polarity.

Mathematically the force F between the charges can be expressed as,

F α Q1Q2/ R 2

Q1Q2------ Product of two charges

R ------- Distance between the two charges

The law also states that the force depends on the medium in which the point charges are located. The effect of medium is introduced in the equation of force as a constant of proportionality denoted as k.

F = k. Q1 Q2/ R2

The constant of proportionality takes into account the effect of medium.

Therefore,

k = 1/ 4 π

= Permittivity of medium in which charges are located. The units of are farads/metre(F/m)

In general, is expressed as,

Permittivity of free space or vacuum

= Relative permittivity or dielectric constant of the medium with respect to free space.

Absolute permittivity

For free space or vacuum the relative permittivity hence

F = 1/ 4 π . Q1Q2/ R2

The value of permittivity of free space is

1/ 36 π x 10 -9 = 8.854 x 10 -12 F/m

k = 1/ 4π = 1/ 4π x 8.854 x 10 -12 = 8.98 x 10 9 = 9 x 10 9 m/F

Hence the Coulombs law can be expressed as,

F = Q1 Q2 / 4π R 2

This the force between the two-point charges located in free space or vacuum.

Q11) Define electric field intensity and also derive its expression?

A11) Consider a point charge Q1 as shown in figure:

Fig: Electric Field Intensity

If any other similar charge Q2 is brought near it Q2 experiences a force. Infact if Q2 is moved around Q1 still Q2 experiences a force as shown in figure.

Thus, there exists a region around a charge in which it exerts force on any other charge. This region where a particular charge exerts a force on any other charge located in that region called electric field of that charge. The electric field of Q1 is shown in figure (b).

The force experienced by the charge Q2 due to Q1 is given by Coulombs law as,

Thus, force per unit charge can be written as:

This force exerted per unit charge is called electric field intensity or electric field strength. It is a vector quantity and is directed along a segment from the charge Q1 to the position of any other charge.

It is denoted as

Another definition of electric field is the force experienced by a unit positive test charge that is Q2 = 1C.

Consider a charge Q1 as shown in figure below. The unit positive charge Q2=1C is placed at distance R from Q1. Then the force acting on Q2 due to Q1 is along the unit vector . As the charge Q2 is unit charge the force exerted on Q2 is nothing but electric field intensity of Q1. Then the force acting on Q2 due to Q1 is along the unit vector . As the charge Q2 is unit charge the force exerted onQ2 is nothing but electric field intensity of Q1 at a point where unit charge is placed.

Fig: Concept of electric field intensity

If a charge Q1 is located at the centre of the spherical coordinate system then unit vector in equation (3) becomes the radial unit vector coming radially outwards from Q1 and the distance R is the radius of the sphere r.

Q12) Define electric flux?

A12) The region in which the influence of an electric charge can be felt is known as electric field. The force experienced by a unit positive charge at a point is called is electric field strength.

E= F/q N/C or V/m

Electric flux density is defined as the amount of flux passes through unit surface area in the space at right angle to the direction of electric field.

E =

Q: charge

r: distance between the object and field

Q13) Define Gauss Law and also prove it?

A13) The surface integral of electrostatic field produced by any source over any closed surface S enclosing a volume V in vacuum i.e., total electric flux over the closed surface S in vacuum is X the total charge () contained inside S i.e.