Resistance and resistivity ( By pak physics)

Opposition offered by the material of conductor in the flow of electric current is called resistance.
Resistance opposes the flow of current through a conductor. Resistance of a conductor is due to the collision of free electrons with the atoms of the conductor.
It is denoted by “R”

(1) Length of conductor
Resistance of a conductor is directly proportional to the length of conductor .
R a L…………….(a)
(2) Area of cross section of conductor
Resistance of a conductor is inversely proportional to area of cross section of conductor.
R a 1 / A……………….(b) Combining (a) and (b)                                                                       R α L/A                                                 R = rL/A Where  r = resistivity of material of conductor RESISTIVITY Resistivity is an electrical property of material . It is defined as the resistance of a material or conductor of 1 cubic meter volume.                                         Or It is the resistance of a conductor of unit length and unit area.                                         Or Resistivity of a conductor is the resistance of 1 meter long conductor whose area of cross section is  I meter square Unit: R = rL/A     r = ohm x m Different materials have different values of resistivity. A very high value of resistivity indicates high electrical resistance

Electric Current (By pak physics notes)

"Electric current is defined as the amount of electric charge  passing through a cross section of a conductor in unit time."

                                                                              OR

"The rate of flow of electric charge through a cross section of a conductor is called Electric Current". 

 Mathematically 

                                          Electric current = electric charge / time

                                                                 I     = Q/t

Unit of electric current is AMPERE.      1 ampere = 1 coulomb / 1 sec
AMPERE
In S.I system unit of electric current is ampere. Ampere is defined as:                       “Current through a conductor will be 1 ampere if one coulomb of electric                              charge passes through any cross section of conductor in 1 second.”                                                                          1 ampere = 1 coulomb / 1 sec
There are two types of current.
ELECTRONIC CURRENT
Electronic current flows from negative to positive terminal.
CONVENTIONAL CURRENT
Direction of conventional current is taken from higher potential to the lower potential.    (www.pak-physicsnotes.blogspot.com)

Capacitance of capacitor (By pak physics notes)

The ability of a capacitor to store electric charge between its plates is its capacity or capacitance.
DEFINITION
The capacitance of a capacitor is defined as:                                           "The ratio of electric charge stored on any one of the                                   plates of capacitor to potential difference between the plates."
Mathematically
C = Q / V
UNIT OF CAPACITANCE
In S.I. system unit of capacitance is                                   Coulomb / volt                                               OR                                           Farad Farad is a large unit therefore in general practice we use small units (1) uF (microfarad) 1 uF = 1x10-6 F (2) uuF (Pico farad) 1 uuF = 1x10-12 F For latest information , free computer courses and high impact notes visit : www.citycollegiate.com
CAPACITANCE OF A PARALLEL PLATE CAPACITOR
Consider a parallel plate capacitor as shown below:
Let  The area of each plate = A The separation between plates = d Medium = air Surface density of charge on each plate = s  The electric field intensity between the plates of capacitor is given by
E = s / eo
Potential difference between the plates of capacitor can be calculated by the following relation
V = Ed Putting the value of "E" V = (s/eo) x d But s = Q/A V = Qd/A.eo AeoV = Qd…………(a)
We know that  The electric charge stored on any one of plate of capacitor is
Q = CV……………(b) Putting the value of Q in (a) AeoV = CVd Cd = Aeo
C = Aeo/d (www.pak-physicsnotes.blogspot.com)

Columb's Law (By pak physics notes)

STATEMENT

The electrostatic force of attraction or repulsion between two point charges is directly proportional      to the product of charges inversely proportional to the square of distance between them.

Explanation

Consider two point charges q1 and q2 at a placed distance of r from each other. Let the electrostatic force between them is F.

According to the first part of the law:

                                                                            F α q₁q₂

According to the second part of the law:                                                                              F α 1/r2 Combining above statements:                                                                              F α q1q2/r2                                                                              F =k q1q2/r2   --------------------(1) Where k is the constant of proportionality. Value of K is equal to 1/4pe0 where eo is permittivity of free space .Its volume is 8.85 x 10-12 c2/Nm2. Thus in S.I. system numerical value of K is 8.98755 x 109 Nm2c-2. Putting the value of K = 1/4pe0 in equation (i)                                                                                                                                F = 1/4pe0 q1q2/r2 FORCE IN THE PRESENCE OF  DIELECTRICMEDIUM If the space between the charges is filled with a non conducting medium or an insulator called "dielectric", it is found that the dielectric reduces the electrostatic force as compared to free space by a factor (er) called DIELECTRIC CONSTANT. It is denoted byer . This factor is also known as RELATIVE PERMITTIVITY. It has different values for different dielectric materials.                     (www.pak-physicsnotes.blogspot.com)

Gauss's Law (By pak physics notes)

Gauss’s law is a quantitative relation which applies to any closed hypothetical surface called Gaussian surface to determine the total flux (Ø) through the surface and the net charge(q) enclosed by the surface.

Statement

"The total electric flux through a closed surface is equal to 1/e_o 

times the total charge enclosed by the surface."

PROOF

Consider a Gaussian surface as shown below which encloses a number of point charges q₁,q₂,q₃…….q_n
Draw imaginary spheres around each charge. Now we make use of the fact that the electric flux through a sphere is q/e_o.

                                                         

Flux due to q₁ will be Ø₁ = q₁/e_o

Flux due to q₂ will be Ø₂ = q₂/e_o

Flux due to q₃ will be Ø₃ = q₃/e_o

Flux due to q_n will be Ø_n = q_n/e_o

	Hence the total flux Øe will be the sum of all flux i.e
	Ø = Ø1 + Ø2+ Ø3 +Ø4 ……….+ Øn
	Ø = q1/eo+ q2/eo+ q3/eo+……….. +qn/e
	Ø = 1/eo(q1+ q2+ q3+……….. +qn)
	Ø = 1 /eo  x ( total charge)
	This shows that the total electric flux through a closed surface regardless of its shape or size is numerically equal to 1 /eo times the total charge enclosed by the surface.      (www.pak-physicsnotes.blogspot.com)

Elstic and inelastic collision (pak physics notes)

Elastic collision
An elastic collision is that in which the momentum of the system as well as kinetic energy of the system before and after collision is conserved.
Inelastic collision
An inelastic collision is that in which the momentum of the system before and after collision is conserved    but the kinetic energy before and after collision is not conserved.
Elastic collision in one diamension

Consider two non-rotating spheres of mass m1 and m2 moving initially along the line joining their centers with velocities u1 and u2 in the same direction. Let u1 is greater than u2. They collide with one another and after having an elastic collision start moving with velocities v1and v2 in the same directions on the same line. Momentum of the system before collision = m1u1 + m2u2 Momentum of the system after collision = m1v1 + m2v2    According to the law of conservation of momentum: m1u1 + m2u2 = m1v1 + m2v2 m1v1 – m1u1 = m2u2 – m2v2 m1(v1 – u1) = m2(u2 – v2) -------(1)    Similarly                                              K.E of the system before collision = ½ m1u12 + ½ m2u22 K.E of the system after collision = ½ m1v12 + ½ m2v22    Since the collision is elastic, so the K.E of the system before and after collision is conserved .    Thus ½ m1v12 + ½ m2v22 = ½ m1u12 + ½ m2u22 ½ (m1v12 + m2v22) = ½ (m1u12 + ½ m2u22 m1v12-m1u12=m2u22-m2v22 m1(v12-u12) = m2(u22-v22) m1(v1+u1) (v1-u1) = m2(u2+v2) (u2-v2) ------- (2)    Dividing equation (2) by equation (1) V1+U1 = U2+V2    From the above equation V1=U2 +V2 -U1_________(a) V2=V1+U1 -U2_________(b)    Putting the value of V2 in equation (1) m1 (v1-u1) =m2 (u2-v2) m1 (v1-u1) =m2{u2-(v1+u1-u2)} m1(v1-u1)=m2{u2-v1-u1+u2} m1(v1-u1)=m2{2u2-v1-u1} m1v1-m1u1=2m2u2-m2v1-m2u1 m1v1+m2v1=m1u1-m2u1+2m2u2 v1(m1+m2)=(m1-m2)u1-2m2u2    In order to obtain V2 putting the value of V1 from equation (a) in equation (i) m1 (v1-u1) = m2(u2-v2) m1(u2+v2-u1-u1)=m2(u2-v2) m1(u2+v2-2u1)=m2(u2-v2) m1u2+m1v2-2m1u1=m2u2-m2v2 m1v2+m2v2=2m1u1+m2u2-m1u2 v2(m1+m2)=2m1u1+(m2-m1)u2                      (www.pak-physicsnotes.blogspot.com)