Q. No.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
A hollow metal sphere of radius \(R\) is uniformly charged. The electric field due to the sphere at a distance \(r\) from the centre:
| 1. | decreases as \(r\) increases for \(r<R\) and for \(r>R\). |
| 2. | increases as \(r\) increases for \(r<R\) and for \(r>R\). |
| 3. | is zero as \(r\) increases for \(r<R\), decreases as \(r\) increases for \(r>R\). |
| 4. | is zero as \(r\) increases for \(r<R\), increases as \(r\) increases for \(r>R\). |
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:
| 1. | \(qEy\) | 2. | \(2qEy\) |
| 3. | 4. |
The figure shows the electric lines of force emerging from a charged body. If the electric field at \(A\) and \(B\) are \(E_A\) and \(E_B\) respectively and if the displacement between \(A\) and \(B\) is \(r,\) then:
1. \(E_A>E_B\)
2. \(E_A<E_B\)
3. \(E_{A} = \frac{E_{B}}{r}\)
4. \(E_{A} = \frac{E_{B}}{r^{2}}\)
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)
| 1. | \(\dfrac{13q}{4\pi \varepsilon_0r^2}\) | 2. | zero |
| 3. | \(\dfrac{q}{4\pi \varepsilon_0r^2}\) | 4. | \(\dfrac{12q}{4\pi \varepsilon_0r^2}\) |
| 1. | \(\dfrac{Eqm}{t}\) | 2. | \(\dfrac{E^2q^2t^2}{2m}\) |
| 3. | \(\dfrac{2E^2t^2}{qm}\) | 4. | \(\dfrac{Eq^2m}{2t^2}\) |
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. \(2v\)
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. | \(1.5~\text{m/s},~ 3~\text{m/s}\) |
| Assertion (A): | Point charges \(q_{1}\) and \(q_{2}\) produce electric field of magnitude \(E_{1}\) and \(E_{2}\) at a point and potential \(V_{1}\) and \(V_{2}\) at the same point. The electric field due to both the charges at that point must be \(E_{1}+E_{2}.\) |
| Reason (R): | The electric potential at that point due to both the charges must be \(V_{1}+V_{2}.\) |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | (A) is False but (R) is True. |
| 1. | zero | 2. | \(4\dfrac{kq}{a^2}\) |
| 3. | \(2\dfrac{kq}{a^2}\) | 4. | \(2\sqrt2\dfrac{kq}{a^2}\) |
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