In relativistic mechanics `m=(m_(0))/(sqrt((1-(v^(2))/(c^(2))))` the equivalent relation in electricity for electric charge is

Relativistic mass is given by m_o/sqrt(1-upsilon^2/-

A boy recalls the relation for relativistic mass (m) in terms of rest mass (m_(0)) velocity of particle v, but forgets to put the constant c (velocity of light). He writes m= (m_(0))/((1-v^(2))^(1//2)) correct the equation by putting the missing 'c'.

The momentum in mechanics is expressed as m xx v . The analogous expression in electricity is

Electric potential in a region is varying according to the relation V = (3x^(2))/(2) - (y^(2))/(4) , where x and y are in metre and V is volt, Electric field intensity (in N/C) at a point (1 m, 2 m ) is

A boy recalls the relativistic mass wrongly as m=m_0/sqrt(1-V^2) Using dimensional method put the missing 'C' at proper place.

Two equal and opposite charges of masses m_(1) and m_(2) are accelerated in a uniform electric field through the same distance. What is the ratio of their accelerations, if their ratio of masses is m_(1)/m_(2)=0.5 ?

Two equal and opposite charges of masses m_(1) and m_(2) are accelerated in a uniform electric field through the same distance. What is the ratio of their accelerations, if their ratio of masses is m_(1)/m_(2)=0.5 ?