Q. No.The acceleration of an electron due to the mutual attraction between the electron and a proton when they are \(1.6~\mathring{A}\) apart is:
( \(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9~ \text{Nm}^2 \text{C}^{-2}\) )
1.
\( 10^{24} ~\text{m/s}^2\)
2
\( 10^{23} ~\text{m/s}^2\)
3.
\( 10^{22}~\text{m/s}^2\)
4.
\( 10^{25} ~\text{m/s}^2\)
( \(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9~ \text{Nm}^2 \text{C}^{-2}\) )
The figure shows electric field lines in which an electric dipole p is placed as shown. Which of the following statements is correct?
| 1. | The dipole will not experience any force. |
| 2. | The dipole will experience a force towards the right. |
| 3. | The dipole will experience a force towards the left. |
| 4. | The dipole will experience a force upwards. |
The electric field at a point on the equatorial plane at a distance \(r\) from the centre of a dipole having dipole moment \(\overrightarrow{P}\) is given by:
(\(r\gg\) separation of two charges forming the dipole, \(\varepsilon_{0} =\) permittivity of free space)
| 1. | \(\overrightarrow{E}=\frac{\overrightarrow{P}}{4\pi \varepsilon _{0}r^{3}}\) | 2. | \(\overrightarrow{E}=\frac{2\overrightarrow{P}}{\pi \varepsilon _{0}r^{3}}\) |
| 3. | \(\overrightarrow{E}=-\frac{\overrightarrow{P}}{4\pi \varepsilon _{0}r^{2}}\) | 4. | \(\overrightarrow{E}=-\frac{\overrightarrow{P}}{4\pi \varepsilon _{0}r^{3}}\) |
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 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 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 \(\frac{E}{2}\) then its terminal velocity will be:
1. \(\frac{v}{2}\)
2. \(v\)
3. \(\frac{3v}{2}\)
4. \(2v\)
Refer to the arrangement of charges in the figure and a Gaussian surface of radius \(R\) with \(Q\) at the centre. Then:
| a. | total flux through the surface of the sphere is \(\frac{-Q}{\varepsilon_0}\). |
| b. | field on the surface of the sphere is \(\frac{-Q}{4\pi \varepsilon_0 R^2}.\) |
| c. | flux through the surface of the sphere due to \(5Q\) is zero. |
| d. | field on the surface of the sphere due to \(-2Q\) is the same everywhere. |
Choose the correct statement(s):
1. (a) and (d)
2. (a) and (c)
3. (b) and (d)
4. (c) and (d)
| a. | on any surface. |
| b. | if the charge is outside the surface. |
| c. | could not be defined. |
| d. | if charges of magnitude \(q\) were inside the surface. |
1. (a) and (d)
2. (a) and (c)
3. (b) and (d)
4. (c) and (d)
Two point dipoles of dipole moment \(\vec{p}_{1}\) and \(\vec{p}_{2}\) are at a distance \(x\) from each other and \(\vec{p}_{1} \left|\right| \vec{p}_{2}\). The force between the dipole is:
1. \(\frac{1}{4 π\varepsilon_{0}} \frac{4 p_{1} p_{2}}{x^{4}}\)
2. \(\frac{1}{4 π\varepsilon_{0}} \frac{3 p_{1} p_{2}}{x^{3}}\)
3. \(\frac{1}{4π\varepsilon_{0}} \frac{6 p_{1} p_{2}}{x^{4}}\)
4. \(\frac{1}{4 π\varepsilon_{0}} \frac{8 p_{1} p_{2}}{x^{4}}\)
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 \(r>R\). |
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