Two charges,each equal to 2C can be placed so that the force between them equals to the weight of an object of mass 1kg .The distance between them is (take "g=10m/s^(2)" ) (10 pi varepsilon_(0))^(1/2), [1/(10 pi varepsilon_(0))^(1/2)], [1/10 pi varepsilon_(0)], [10 pi varepsilon_(0)]

At what separation should two equal charges 1.0 C each, be placed so that the force between them equals the weight of a 50 kg person?

Two bodies each of mass 10 kg attract each other with a force of 0.1 milligram weight. What is the distance between them ?

Two equal charges are placed at a separtion of 1.0 m. What should be the magnitude of the charges so that the force between them equals of a 50 kg person?

Three charges -q +q and +q are placed at the corner of the equilateral triangle of side a as shown in the figure.The net electric potential at the centroid of the triangle will be (k=(1)/(4 pi varepsilon_(0)))

A charge q_(1) is placed at the centre of a spherical conducting shell of radius "R" .Conducting shell has a total charge q_(2) .Electrostatic potential energy of the system is (A) (q_(1)^(2)+2q_(1)q_(2))/(8 pi varepsilon_(0)R) (B) (q_(2)^(2)+2q_(1)q_(2))/(8 pi varepsilon_(0)R) (C) (q_(1)^(2)+q_(1)q_(2))/(8 pi varepsilon_(0)R) (D) (q_(2)^(2)+q_(1)q_(2))/(8 pi varepsilon_(0)R)

If (pi)/(2)

A point charge q having mass m is placed at rest at point A at distance r from a short dipole of dipole moment P as shown in the figure.The initial acceleration of charge is a=(nkPq)/(2mr^(3)) (k=(1)/(4 pi varepsilon_(0))) . Value of n is ?

If 2f(x^(2))+3f((1)/(x^(2)))=x^(2)-1x varepsilon R_(0) then f(x^(2)) is

Calculate the gravitational force of attraction between two spherical bodies, each of mass 1kg placed at 10m apart (G = 6.67 xx 10^(-11)Nm^(2)//kg^(2)) .