The electric field in a certain region is acting radially outward and is given by \(E=Aa.\) A charge contained in a sphere of radius \(a\) centered at the origin of the field will be given by:
1. \(4 \pi \varepsilon_{{o}} {A}{a}^2\)
2. \(\varepsilon_{{o}} {A} {a}^2\)
3. \(4 \pi \varepsilon_{{o}} {A} {a}^3\)
4. \(\varepsilon_{{o}} {A}{a}^3\)

Subtopic:  Gauss's Law |
 68%
From NCERT
NEET - 2015
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Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is \(r\) (as shown in Fig. I). Now, as shown in Fig. II, the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes:
     
1. \(\frac{r}{\sqrt[3]{2}}\)
2. \(\frac{r}{\sqrt[2]{2}}\)
3. \(\frac{2r}{3}\)
4. none of the above

Subtopic:  Coulomb's Law |
 70%
From NCERT
AIPMT - 2013
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An electric dipole of dipole moment \({p}\) is aligned parallel to a uniform electric field \({E}.\) The energy required to rotate the dipole by \(90^\circ\) is:
1. \( p^2 E\)
2. \( p E\)
3. infinite
4. \( {pE}^2\)
Subtopic:  Electric Dipole |
 90%
From NCERT
NEET - 2013
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A charge \(q\) is placed at the centre of the line joining two equal positive charges \(Q.\) The system of the three charges will be in equilibrium, if \(q\) is equal to:
1. \(\dfrac{-Q}{4}\) 2. \(\dfrac{Q}{4}\)
3. \(\dfrac{-Q}{2}\) 4. \(\dfrac{Q}{2}\)
 
Subtopic:  Coulomb's Law |
 68%
From NCERT
NEET - 2013
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What is the flux through a cube of side \(a,\) if a point charge of \(q\) is placed at one of its corners?
1. \(\frac{2q}{\varepsilon_0}\)
2. \(\frac{q}{8\varepsilon_0}\)
3. \(\frac{q}{\varepsilon_0}\)
4. \(\frac{q}{2\varepsilon_0}\)

Subtopic:  Gauss's Law |
 88%
From NCERT
AIPMT - 2012
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A charge \(Q\) is enclosed by a Gaussian spherical surface of radius \(R\). If the radius is doubled, then the outward electric flux will:
1. be reduced to half
2. remain the same
3. be doubled
4. increase four times
Subtopic:  Gauss's Law |
 88%
From NCERT
AIPMT - 2011
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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 |
 78%
From NCERT
AIPMT - 2010
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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 |
 76%
From NCERT
AIPMT - 2010
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The electric field at a distance \(\frac{3R}{2}\) from the centre of a charged conducting spherical shell of radius \(R\) is \(E\). The electric field at a distance \(\frac{R}{2}\) from the centre of the sphere is:
1. \(E\)
2. \(\frac{E}{2}\)
3. \(\frac{E}{3}\)
4. zero
 

Subtopic:  Gauss's Law |
 85%
From NCERT
AIPMT - 2010
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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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