A uniform electric field `E_0` is directed along positive y-driection. Find the change in electric potential energy of a positive test charge `q_0` when it is displaced in this field from `y_i=a` to `y_f=2a` along the y-axis.

A uniform electric field of 400 V//m is directed at 45^(@) to the positive X-axis, in the X-Y plane as shown in figure

A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge.

A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge.

A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge.

The uniform electric field is along positive y-axis positive x axis as shown in figure. The signs of potential differences V_P-V_Q and V_B-V_A respectively are

An electron is projected along positive z-axis, magnetic field is present along y-axis then direction of initial acceleration of the particle is

There exists a uniform electric field E =4 xx 10^(5) Vm^(-3) directed along negative x-axis such that electric potential at origin is zero. A charge of - 200 mu C is placed at origin and a charged of + 200 mu C is placed at a (3 m, 0) . The electrostatic potential energy of the system is

A positively charged ring is in y - z plane with its centre at origin . A positive test charge q_(e ) , held at origin is released along x axis, then its speed

Uniform electric field of magnitude 100 Vm^-1 in space is directed along the line y = 3 + x . Find the potential difference between points A (3,1) and B (1,3) .

An electric dipole has the magnitude of its charge as q and its dipole moment is p . It is placed in a uniform electric field E . If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively (a) 2qE and minimum (b) qE and pE (c) zero and minimum (d) qE and maximum