If the electric potential at any point (x, y, z) m in space is given by V = 3x² volt. The electric field at the point (1, 0, 3) m will be :

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 $$\times$$ 10⁵ NC^$$-$$1. If the charge on the particle is 40 $$\mu$$C and the initial velocity is 200 ms^$$-$$1, how much distance it will travel before coming to the rest momentarily :

Two point charges A and B of magnitude +8 $$\times$$ 10^$$-$$6 C and $$-$$8 $$\times$$ 10^$$-$$6 C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 $$\times$$ 10⁴ NC^$$-$$1. The distance 'd' between the point charges A and B is :

Given below are two statements :

Statement I : A point charge is brought in an electric field. The value of electric field at a point near to the charge may increase if the charge is positive.

Statement II : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero.

Choose the correct answer from the options given below :