Twelve point charges each of charge \(q\) C are placed at the circumference of a circle of radius \(r\) m with equal angular spacing. If one of the charges is removed, the net electric field (in N/C) at the centre of the circle is:
(\(\varepsilon_0 \)-permittivity of free space)
1. \(\frac{13q}{4\pi \varepsilon_0r^2}\)
2. zero
3. \(\frac{q}{4\pi \varepsilon_0r^2}\)
4. \(\frac{12q}{4\pi \varepsilon_0r^2}\)

A spherical conductor of radius \(10~\text{cm}\) has a charge of \(3.2 \times 10^{-7}~\text{C}\) distributed uniformly. What is the magnitude of the electric field at a point \(15~\text{cm}\) from the center of the sphere? \(\frac{1}{4\pi \varepsilon _0} = 9\times 10^9~\text{N-m}^2/\text{C}^2\)
1. \(1.28\times 10^{5}~\text{N/C}\)
2. \(1.28\times 10^{6}~\text{N/C}\)
3. \(1.28\times 10^{7}~\text{N/C}\)
4. \(1.28\times 10^{4}~\text{N/C}\)

A hollow metal sphere of radius \(R\) is uniformly charged. The electric field due to the sphere at a distance \(r\) from the centre:

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:

A toy car with charge \(q\) moves on a frictionless horizontal plane surface under the influence of a uniform electric field \(\vec E.\)Due to the force \(q\vec E,\) its velocity increases from \(0\) to \(6~\text{m/s}\) in a one-second duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between \(0\) to \(3\) seconds are respectively:
1. \(2~\text{m/s}, ~4~\text{m/s}\)
2. \(1~\text{m/s}, ~3~\text{m/s}\)
3. \(1~\text{m/s}, ~3.5~\text{m/s}\)
4. \(1.5~\text{m/s},~ 3~\text{m/s}\)

A square surface of side \(L\) (metre) in the plane of the paper is placed in a uniform electric field \(E\) (volt/m) acting along the same plane at an angle θ with the horizontal side of the square as shown in the figure. The electric flux linked to the surface in the unit of V-m is:
     

A thin conducting ring of radius \(R\) is given a charge \(+Q.\) The electric field at the centre O of the ring due to the charge on the part AKB of the ring is \(E.\) The electric field at the centre due to the charge on the part ACDB of the ring is:

                      

The electric field at centre \(O\) of a semicircle of radius \(a\) having linear charge density \(\lambda\) is given by:

     
1. \(\frac{2\lambda}{\epsilon_0 a}\)
2. \(\frac{\lambda\pi}{\epsilon_0 a}\)
3. \(\frac{\lambda}{2\pi \epsilon_0 a}\)
4. \(\frac{\lambda}{\pi \epsilon_0 a}\)
 

In 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 upward direction. If electric field reduces to E/2 then its terminal velocity will be:
1. v/2
2. v
3. 3v/2
4. 2v