Charge and Coulomb’s Law |आवेश तथा कूलाम का नियम | Electrostatic Part 1 | NCERT Class 12

Evolution part 1|class 12

Column I shows charge distribution and column II shows electrostatic field created by that charge at a point r distance from its centre. {:(,"Column I ",,"Column II"),((A),"A stationary point charge",(p),E prop r^(0)),((B),"A stationary uniformly long charge rod",(q),Eprop r^(1)),((C ), "A stationary electric dipole",(r ),E prop r^(*-1)),,(s),E prop r^(-2) ),(,,(t ),E propr^(-3)):}

Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force between two stationary point masses, both have inverse square dependence on the distance between the charges // masees (a) compare the strength of these forces by determining the ratio of their maagnitudes (i) for an electron and as proton (ii) for two protons (b) estimate the accelerations for election and proton due to electrical force of their mutal attraction when they are 1 A apart.

The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the forces between neutron and proton in deuteron as given in the form of a Coulomb force but with an effective charge e': F=1/(4piepsilon_(0)) (e'^(2))/r^(2) estimate the value of (e'//e) given that the following binding energy of a deuteron is 2.2MeV.

The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the forces between neutron and proton in deuteron as given in the form of a Coulomb force but with an effective charge e': F=1/(4piepsilon_(0)) (e'^(2))/r^(2) estimate the value of (e'//e) given that the following binding energy of a deuteron is 2.2MeV.

Two charges Q_(1) and Q_(2) Coulombs are shown in fig. A third charge Q_(3) coulomb is moved from point R to S along a circular path with P as centre Change in potential energy is

Three capacitors C_(1)=6muF,C_(2)=12muF and C_(3)=20muF are connected to a 100 V battery, as shown in figure below: (i) Charge on each plate of capacitor C_(1) . (ii) Electrostatic potential energy stored in capacitor C_(3) .

S_1 : When a positively charged particle is released in an electric field , in its subsequent motion,it may or may not move along the electric field line passing through the point it has been released. S_2 : In electrostatic, conductors are always equipotential surfaces. S_3 : The flux through a closed Gaussian surface is non-zero. The electric field at some point on the Gaussian surface may be zero.

Following experiment was performed by J.J. Thomson in order to measure ratio of charge e and mass m of electron. Electrons emitted from a hot filament are accelerated by a potential difference V. As the electrons pass through deflecting plates, they encounter both electric and magnetic fields. the entire region in which electrons leave the plates they enters a field free region that extends to fluorescent screen. The entire region in which electrons travel is evacuated. Firstly, electric and magnetic fields were made zero and position of undeflected electron beam on the screen was noted. The electric field was turned on and resulting deflection was noted. Deflection is given by d_(1) = (eEL^(2))/(2mV^(2)) where L = length of deflecting plate and v = speed of electron. In second part of experiment, magnetic field was adjusted so as to exactly cancel the electric force leaving the electron beam undeflected. This gives eE = evB . Using expression for d_(1) we can find out (e)/(m) = (2d_(1)E)/(B^(2)L^(2)) A beam of electron with velocity 3 xx 10^(7) m s^(-1) is deflected 2 mm while passing through 10 cm in an electric field of 1800 V//m perpendicular to its path. e//m for electron is