A charged particle \(q\) of mass \(m\) is released on the \(y\text-\)axis at \(y=a\) in an electric field \(\vec E = -4y \hat{j}.\) The speed of the particle on reaching the origin will be:
1. \(\sqrt{\frac{2 a}{m q}}\)
2. \(\frac{a}{\sqrt{m q}}\)
3. \(2 a \sqrt{\frac{q}{m}}\)
4. \(2 \sqrt{\frac{a}{m q}}\)

A charge \(q\) is placed in a uniform electric field \(E.\) If it is released, then the kinetic energy of the charge after travelling distance \(y\) will be:

An infinite number of electric charges each equal to \(5\) nC (magnitude) are placed along the \(x\text-\)axis at \(x=1\) cm, \(x=2\) cm, \(x=4\) cm, \(x=8\) cm ………. and so on. In the setup if the consecutive charges have opposite sign, then the electric field in Newton/Coulomb at \(x=0\) is: \(\left(\frac{1}{4 \pi \varepsilon_{0}} = 9 \times10^{9} ~\text{N-m}^{2}/\text{C}^{2}\right)\)
1. \(12\times 10^{4}\)
2. \(24\times 10^{4}\)
3. \(36\times 10^{4}\)
4. \(48\times 10^{4}\)

In the following four situations, charged particles are at an equal distance from the origin. Arrange the magnitude of the net electric field at origin, starting with the highest.

(i)	(ii)
(iii)	(iv)

Twelve point charges each of charge \(q~\text C\) are placed at the circumference of a circle of radius \(r~\text{m}\) with equal angular spacing. If one of the charges is removed, the net electric field (in \(\text{N/C}\)) at the centre of the circle is:
(\(\varepsilon_0\text- \)permittivity of free space)

An electron falls from rest through a vertical distance \(h\) in a uniform and vertically upward-directed electric field \(E.\) The direction of the electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest through the same vertical distance \(h.\) The fall time of the electron in comparison to the fall time of the proton is:
1. smaller
2.  \(5\) times greater
3.  \(10\) times greater
4. equal

In the Millikan oil drop experiment, a charged drop falls with a terminal velocity \(v.\) If an electric field \(E\) is applied vertically upwards it moves with terminal velocity \(2v\) in the upward direction. If the electric field reduces to \(\frac{E}{2}\) then its terminal velocity will be:
1. \(\frac{v}{2}\)
2. \(v\)
3. \(\frac{3v}{2}\)
4. \(2v\)

Three identical positive point charges, as shown are placed at the vertices of an isosceles right-angled triangle. Which of the numbered vectors coincides in direction with the electric field at the mid-point \(M\) of the hypotenuse?
                 
1. \(1\)
2. \(2\)
3. \(3\)
4. \(4\)

An electron enters an electric field with its velocity in the direction of the electric lines of force. Then: