SAI'S SCIENCE AND MATHS CLASSES

Calculate the work done in moving a charge q from i) A to B and from ii) A to C as shown in the figure, The radius of the sphere is R and point B is at a distance of r from the centre O of the sphere and A and C are   points inside the sphere, the Charge enclosed by the sphere is Q Let the potential at A is V A and that of B is V B the work done is W AB =q(V B -V A ) Note: 1) Potential inside the sphere is a constant= potential on the surface of the sphere            2)Since electric field E=dv/dl =d(constant)/dl=0 ie field inside the sphere is zero            3) Thus E=0 inside a charged sphere and is known as Electrstatic shielding i) V A =(1/4πε 0 ) Q/R    V B =(1/4πε 0 ) Q/r W AB =q (V B -V A ) W AB =( 1/4πε 0 ) Qq[(1/r)-(1/R)] • • • W AB =( Q q / 4πε 0 ) [( R-r )/ Rr ] ii) W AC = q(V C -V A ) inside the sphere the potential is the same ie V C =V A • • • W AC =0 Prepared by Sai’s Classes Poonkunnam Thrissur +919846088761

Find the ratio of Kinetic energy ,of a proton and   an electron when it is approaching from rest towards a Uniform electric field of strength E, after t seconds The force acting on the charge q, F=q E F=ma= q E-------------(1) a=(v-u)/t a=v / t   (since   u=0) v=at KE=½ mv 2 a=qE/m (From (1) ) v=qEt/m KE=½ m (qEt/m) 2 KE=½(qEt) 2 /m ( Note the formula for competitive exams) KE p =½(eEt) 2 /m p KE e =½(eEt) 2 /m e KE p / KE e = m e / m p =1 : 1833 Prepared by Sai’s Classes Poonkunnam Thrissur +919846088761

In an adiabatic process the following observations are obtained Find the value of γ of the working fluid Pressure 20 bar 10 bar Volume 5 lit 7 lit For an adiabatic process PV γ =Constant 20×5 γ = 10×7 γ 2=(7/5) γ taking logaritham on both sides log2= γ (log7-log5) γ = 0.3010/(0.8451-0.6990) γ= 0.3010 / 0.1461 = 2.06 γ=2.06 Prepared by Sai’s Classes Poonkunnam Thrissur +919846088761