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 .

Three charges of +0.1 C each are placed at the vertices of an equilateral triangle of each side 1m. If the energy is supplied at the rate of 1.0 kw, how many hours would be required to move one of the charges on to the mid point of the line joining the other two? Given =1/(4piepsilon_0) = 9xx 10^9 N ^2C^-2

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

An alpha particle having 12.5 MeV kinetic energy collides with copper nucleus (Z=29) and retraces it s path .What is the size of the nucleus? Given, 1/(4 pi e_0) = 9 xx 10^9 N m^2 C^(-2) .

What will be the corresponding expression for the energy needed to completely diassemble the planet earth against the gravitational pull amongst its constituent particles? Assume the earth to be a sphere of uniform mass density. calculate the energy, given that the product of the mass and the radius of the earth to be 2.5xx10^(31)kg-m

Calculate the radius of oxygen nucleus .Given that R_0 = 1.1 xx 10^(-15) m .

A charge of 4xx10^-9 C is distributed uniformly over the circumference of a conducting ring of radius 0.3 m. Calculate the field intensity at a point on the axis of the ring at 0.4 m from its centre. Also, calculate the electri field at the centre of the ring.

A drop of water of mass 18xx10^-3 g falls away from the bottom of a charged conducting sphere of radius 20 cm, carrying with it a charge of 10^(-9)C and leaving on the sphere a uniformly distributed charge of 2.5xx10^(-6)C . What is the speed of the drop after it has fallen 30 cm? (4piepsi_(0))^(-1)=9xx10^(9)JmC^(-2)

Calculate the escape velocity for a particle 1600km above earth's surface. Given g= 9.8 m//s^(2), R= 6400km

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).