We define the electric field strength as:
`vecE = underset (q_0 rarr0)lim vecF/q_0`
Keeping in mind that minimum value of the charge that exists in nature cannot be less than `e = 1.6 xx 10^-19 C`, how can we justify the above definition?

The electric field E due to any point charge is defined as E=underset (q rarr 0)lim F/q , while q is the test charge and F is the force acting on it. What is the physical signification of underset (q rarr 0)lim in this expression?

A positivity charges ball hangs from a silk thread Electric filed at a certain point (at the same horizontal level of ball) due to this charge is E.. We put a positive test charge Q_0 at a point and measure F//q_0 then, it can be prediced that the electric field strength E

In defining electric field due to a point charge, the test charge has to be vanishingly small. How this condition can be justified, when we know that charge less than the on an electron or a proton is not possible?

There is another useful system of units, besides the SI/MKS. A system, called the cgs system. In this system Coulumb's law is given by vecF = (Qq)/(r^2) hatr where the distance r is measured in cm (=10^(-2) m) , F in dynes (= 10^(-5) N) and the charges in electrostatic units (es units), where 1es unit of charge = 1/[3] xx 10^(-9) C . The number [3] actually arises the speed of light in vacuum which now taken to be exactly given by c = 2.99792458 xx 10^8 m/s . An approximate value of c then is c = [3] xx 10^8 m/s . Write 1 esu of charge = x C, where x is a dimensoinless number. Show that this gives 1/(4pi epsilon_0) = (10^(-9)/x^2)(N.m^2)/(C^2) with x = 1/[3] xx 10^(-9) , we have 1/(4pi epsilon_0)= [3]^2 xx 10^9 (Nm^2)/(C^2) or 1/(4pi epsilon_0)= (2.99792458)^2 xx 10^9 (Nm)^2/(C)^2 (exactly).

An electron is circulating around the nucleus of a hydrogen ato in a circular orbit of radius 5.3 cxx10^-11 m. Calculate the electric potential energy of the atom in eV. What would be the electric potential due to a helium nucleus at the same radius. Given, that (4piepsilon_0)^-1 = 9xx10^9 m F^-1 and e = 1.6 xx 10^-19 C .

An electron after bein accelerated through a potential difference of 1 00 V enters a uniform magnetic field of 0.004 T perpendiular to its direction of motion. Calculate the radius of the path describe by the electron. Given m = 9.1 xx 10^-31 kg and e = 1.6 xx 10^-19 C

classically an electron can be in any orbit around nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? Thequestion had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as followswith the basic constants of nature e, me, c and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10 m).- You will find that the length obtained above is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for else h had already made its appearance elsewhere. Bohr lay in recognising that h, m_e , and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.

Shown in the figure four point charges at the corners of a square of side 2 cm. Find the magnitude and direction of the electric field at the centre O of the square, if Q = 0.02 muC Use 1/(4piepsilon_0) = 9 xx 10^9 N m^2 C^-2

A right circular cylinder of length L (in m) and of radius R (in m) has its center at the origin and its length along the X-axis. A uniform electric field vecE = E_x hat I N C^-1 for x > 0 and vec E = -E_xhati NC^-1 for x < 0 exists over the cylinder. Find the net charge enclosed by the cylinder.