Electrostatics ...

The electric potential V is given as a function of distance x (metre) by V = (5x² + 10x – 9) volt. Value of electric field at x =1 is

Two metal pieces having a potential difference of 800V are 0.02 m apart horizontally. A particle of mass 1.96 × 10^–15 kg is suspended in equilibrium between the plates. If e is the elementary charge, then charge on the particle is

The figure shows some of the electric field lines corresponding to an electric fields. The figure suggests

A hollow insulated conducting sphere is given a positive charge of 10µC. What will be the electric field at the centre of the sphere if its radius is 2 metres?

An electron of mass m_e initially at rest moves through a certain distance in a uniform electric field in time t₁. A proton of mass m_p also initially at rest takes time t₂ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio of t₂ / t₁ is nearly equal to

A charged water drop whose radius is 0.1 µm is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be (g = 10m^–1)

Four charges are placed on corners of a square as shown in figure having side of 5 cm If Q is one micro coulomb, then electric field intensity at centre will be

A sphere of radius 1 cm has potential of 8000 V, then energy density near its surface will be

Point charges +4q, –q and +4q are kept on the x-axis at points x = 0, x = a and x= 2a respectivley, then

Two point charges of 20µC and 80µC are 10 cm apart. Where will the electric field strength be zero on the line joining the charges form 20µC charge?