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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. $$10^{25} ~\text{m/s}^2$$

Subtopic:  Coulomb's Law |
75%
From NCERT
NEET - 2020
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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.

Subtopic:  Electric Dipole |
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The electric field at a point on the equatorial plane at a distance $$r$$ from the centre of a dipole having dipole moment $\stackrel{}{}$$$\overrightarrow{P}$$ is given by:
($$r\gg$$ separation of two charges forming the dipole, $$\epsilon_{0} =$$ permittivity of free space)
1. $$\overrightarrow{E}=\frac{\overrightarrow{P}}{4\pi \epsilon _{0}r^{3}}$$

2. $$\overrightarrow{E}=\frac{2\overrightarrow{P}}{\pi \epsilon _{0}r^{3}}$$

3. $$\overrightarrow{E}=-\frac{\overrightarrow{P}}{4\pi \epsilon _{0}r^{2}}$$

4. $$\overrightarrow{E}=-\frac{\overrightarrow{P}}{4\pi \epsilon _{0}r^{3}}$$

Subtopic:  Electric Dipole |
62%
From NCERT
NEET - 2020
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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. $\frac{qEy}{2}$
4. $\sqrt{qEy}$

Subtopic:  Electric Field |
75%
From NCERT
AIPMT - 1998
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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 $$E$$
Subtopic:  Electric Dipole |
66%
From NCERT
AIPMT - 1998
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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

Subtopic:  Electric Field |
From NCERT
AIPMT - 1999
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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 $\frac{-\mathrm{Q}}{{\epsilon }_{0}}$. b. field on the surface of the sphere is $\frac{-Q}{4\pi {\epsilon }_{0}{R}^{2}}$. 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

Subtopic:  Gauss's Law |
71%
From NCERT
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If there were only one type of charge in the universe, then,

 1 ${\oint }_{s}E.dS\ne 0$ on any surface. 2 ${\oint }_{s}E.dS=0$ if the charge is outside the surface. 3 ${\oint }_{s}E.dS$ could not be defined. 4 ${\oint }_{s}E.\mathrm{dS}=\frac{q}{{\epsilon }_{0}}$ if charges of magnitude q were inside the surface.
Choose the correct statement(s):
1. a and d
2. a and c
3. b and d
4. c and d
Subtopic:  Gauss's Law |
72%
From NCERT
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Two point dipoles of dipole moment ${\stackrel{\to }{\mathrm{p}}}_{1}$ and ${\stackrel{\to }{\mathrm{p}}}_{2}$ are at a distance x from each other and ${\stackrel{\to }{\mathrm{p}}}_{1}||{\stackrel{\to }{\mathrm{p}}}_{2}$. The force between the dipole is:

1. $\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{4{p}_{1}{p}_{2}}{{x}^{4}}$

2. $\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{3{p}_{1}{p}_{2}}{{x}^{3}}$

3. $\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{6{p}_{1}{p}_{2}}{{x}^{4}}$

4. $\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{8{p}_{1}{p}_{2}}{{x}^{4}}$

Subtopic:  Electric Dipole |
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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 $$rR$$. 2 increases as $$r$$ increases for $$rR$$. 3 is zero as $$r$$ increases for $$rR$$. 4 is zero as $$r$$ increases for $$rR$$.
Subtopic:  Gauss's Law |
81%
From NCERT
NEET - 2019
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