Q. No.| 1. | \(60\times10^{3}~\text{Vm}^{-1}\) | 2. | \(90\times10^{3}~\text{Vm}^{-1}\) |
| 3. | zero | 4. | infinite |
A dipole with moment \(\vec p\) is placed in a uniform electric field \(\vec E\). The torque acting on the dipole is given by:
1. \(\vec{\tau }=\vec{p}\cdot \vec{E}\)
2. \(\vec{\tau }=\vec{p} \times \vec{E}\)
3. \(\vec{\tau }=\vec{p}+ \vec{E}\)
4. \(\vec{\tau }=\vec{p}- \vec{E} \)
| 1. | \(\dfrac{Q}{6\varepsilon_0}\times10^{-6}\) | 2. | \(\dfrac{Q}{6\varepsilon_0}\times10^{-3}\) |
| 3. | \(\dfrac{Q}{2\varepsilon_0}\) | 4. | \(\dfrac{Q}{8\varepsilon_0}\) |
A point \(Q\) lies on the perpendicular bisector of an electric dipole of dipole moment \(p.\) If the distance of \(Q\) from the dipole is \(r\) (much larger than the size of the dipole), then the electric field at \(Q\) is proportional to:
1. \(p^{2}\) and \(r^{-3}\)
2. \(p\) and \(r^{-2}\)
3. \(p^{-1}\) and \(r^{-2}\)
4. \(p\) and \(r^{-3}\)
The electric field at centre \(O\) of a semicircle of radius \(a\) having linear charge density \(\lambda\) is given by:
| 1. | \(\dfrac{2\lambda}{\epsilon_0 a}\) | 2. | \(\dfrac{\lambda\pi}{\epsilon_0 a}\) |
| 3. | \(\dfrac{\lambda}{2\pi \epsilon_0 a}\) | 4. | \(\dfrac{\lambda}{\pi \epsilon_0 a}\) |
Who evaluated the mass of electron indirectly with help of charge:
1. Thomson
2. Millikan
3. Rutherford
4. Newton
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. |
If a charge \(Q\) is situated at the corner of a cube, the electric flux passing through all six faces of the cube is:
| 1. | \(\frac{Q}{6\varepsilon_0}\) | 2. | \(\frac{Q}{8\varepsilon_0}\) |
| 3. | \(\frac{Q}{\varepsilon_0}\) | 4. | \(\frac{Q}{2\varepsilon_0}\) |
The electric field at the equator of a dipole is \(E.\) If the strength of the dipole and distance are now doubled, then the electric field will be:
| 1. | \(E/2\) | 2. | \(E/8\) |
| 3. | \(E/4\) | 4. | \(E\) |
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. \(2v\)
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