Electrostatics

A charge $$Q$$ is placed at each of the opposite corners of a square. A charge $$q$$ is placed at each of the other two corners. If the net electrical force on $$Q$$ is zero, then $$Q/q$$ equals:

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $${30^ \circ }$$ with each other. When suspended in a liquid of density $$0.8g$$ $$c{m^{ - 3}},$$ the angle remains the same. If density of the material of the sphere is $$1.6$$ $$g$$ $$c{m^{ - 3}},$$ the dielectric constant of the liquid is

Two identical charged spheres suspended from a common point by two massless strings of length $$l$$ are initially a distance $$d\left( {d < < 1} \right)$$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity $$v$$. Then as a function of distance $$x$$ between them,

Two charges, each equals to $$q,$$ are kept at $$x=-a$$ and $$x=a$$ on the $$x$$-axis. A particle of mass $$m$$ and charge $${q_0} = {q \over 2}$$ is placed at the origin. If charge $${q_0}$$ is given a small displacement $$\left( {y < < a} \right)$$ along the $$y$$-axis, the net force acting on the particle is proportional to

Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameters, and the force between theis F. A third identical conducting sphere, C, is uncharged. Sphere C is first touhed to A, then to B, and then removed. As a result, the force between A and B would be equal to :

Three charges + Q, q, + Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by + Q, placed at x = 0, is zero, then value of q is :

Charge is distributed within a sphere of radius R with a volume charge density $$\rho \left( r \right) = {A \over {{r^2}}}{e^{ - {{2r} \over s}}},$$ where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :

Two electrons each are fixed at a distance '2d'. A third charge proton placed at the midpoint is displaced slightly by a distance x (x << d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency : (m = mass of charged particle)

A certain charge Q is divided into two parts q and (Q $$-$$ q). How should the charges Q and q be divided so that q and (Q $$-$$ q) placed at a certain distance apart experience maximum electrostatic repulsion?

Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle '$$\theta$$' with the vertical?