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# The mean free path of electrons in a metal is $$4\times 10^{-8}~\text{m}$$. The electric field which can give an average of $$2~\text{eV}$$ energy to an electron in the metal will be in units of Vm-1: 1. $$8\times 10^{7}$$ 2. $$5\times 10^{-11}$$ 3. $$8\times 10^{-11}$$ 4. $$5\times 10^{7}$$

Subtopic:  Electric Field |
From NCERT
AIPMT - 2009
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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:

1. $$3E$$ along $$KO$$
2. $$E$$ along $$OK$$
3. $$E$$ along $$KO$$
4. $$3E$$ along $$OK$$

Subtopic:  Electric Field |
75%
From NCERT
AIPMT - 2008
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Three-point charges $$+q$$, $$-2q$$ and $$+q$$ are placed at points $$(x=0,y=a,z=0)$$$$(x=0, y=0,z=0)$$ and $$(x=a, y=0, z=0)$$, respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are:

 1 $$\sqrt{2}qa$$ along $$+y$$ direction 2 $$\sqrt{2}qa$$ along the line joining points $$(x=0,y=0,z=0)$$ and $$(x=a,y=a,z=0)$$ 3 $$qa$$ along the line joining points $$(x=0,y=0,z=0)$$ and $$(x=a,y=a,z=0)$$ 4 $$\sqrt{2}qa$$ along $$+x$$ direction
Subtopic:  Electric Dipole |
84%
From NCERT
AIPMT - 2007
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A hollow cylinder has a charge $$q$$ coulomb within it (at the geometrical centre). If $$\phi$$ is the electric flux in units of Volt-meter associated with the curved surface $$B,$$ the flux linked with the plane surface $$A$$ in units of volt-meter will be:

1. $$\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)$$
2. $$\frac{q}{2\varepsilon_0}$$
3. $$\frac{\phi}{3}$$
4. $$\frac{q}{\varepsilon_0}-\phi$$

Subtopic:  Gauss's Law |
74%
From NCERT
AIPMT - 2007
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A square surface of side $$L$$ (m) is in the plane of the paper. A uniform electric field $$\vec{E}$$ (V/m), also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is:

 1 $$EL^2/ ( 2ε_0 )$$ 2 $$EL^2 / 2$$ 3 zero 4 $$EL^2$$
Subtopic:  Gauss's Law |
81%
From NCERT
AIPMT - 2006
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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 zero as $$r$$ increases for $$rR$$. 4 zero as $$r$$ increases for $$rR$$.
Subtopic:  Electric Field |
76%
From NCERT
NEET - 2019
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Two point charges $$A$$ and $$B$$, having charges $$+Q$$ and $$-Q$$ respectively, are placed at a certain distance apart and the force acting between them is $$F.$$ If $$25\%$$ charge of $$A$$ is transferred to $$B$$, then the force between the charges becomes:

 1 $$\dfrac{4F}{3}$$ 2 $$F$$ 3 $$\dfrac{9F}{16}$$ 4 $$\dfrac{16F}{9}$$
Subtopic:  Coulomb's Law |
76%
From NCERT
NEET - 2019
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Two parallel infinite line charges with linear charge densities $$+\lambda$$ C/m and $$+\lambda$$ C/m are placed at a distance $${R}.$$ The electric field mid-way between the two line charges is:
1. $$\dfrac{\lambda}{2 \pi \varepsilon_0 {R}}$$ N/C
2. zero
3. $$\dfrac{2\lambda}{ \pi \varepsilon_0 {R}}$$ N/C
4. $$\dfrac{\lambda}{ \pi \varepsilon_0 {R}}$$ N/C

Subtopic:  Gauss's Law |
64%
From NCERT
NEET - 2019
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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:

 1 $$EL^{2}$$ 2 $$EL^{2} cos\theta$$ 3 $$EL^{2} sin\theta$$ 4 $$0$$
Subtopic:  Electric Field |
75%
From NCERT
AIPMT - 2010
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Two positive ions, each carrying a charge $$q$$, are separated by a distance $$d$$. If $$F$$ is the force of repulsion between the ions, the number of electrons missing from each ion will be:
($$e$$ is the charge on an electron)

 1 $$\frac{4 \pi \varepsilon_{0} F d^{2}}{e^{2}}$$ 2 $$\sqrt{\frac{4 \pi \varepsilon_{0} F e^{2}}{d^{2}}}$$ 3 $$\sqrt{\frac{4 \pi \varepsilon_{0} F d^{2}}{e^{2}}}$$ 4 $$\frac{4 \pi \varepsilon_{0} F d^{2}}{q^{2}}$$
Subtopic:  Coulomb's Law |
77%
From NCERT
AIPMT - 2010
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