Consider the charge configuration and spherical Gaussian surface as shown in the figure. While calculating the flux of the electric field over the spherical surface, the electric field will be due to: 

(1) q2 only

(2) Only the positive charges

(3) All the charges

(4) +q1 and – q1 only

Subtopic:  Gauss's Law |
 61%
From NCERT
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Two-point charges +q and –q are held fixed at (–d, 0) and (d, 0) respectively of a (x, y) coordinate system. Then 

(1) E at all points on the y-axis is along i^

(2) The electric field E at all points on the x-axis has the same direction

(3) Dipole moment is 2qd directed along i^

(4) Work has to be done in bringing a test charge from infinity to the origin

Subtopic:  Electric Dipole |
From NCERT
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The electric field due to a uniformly charged solid sphere of radius R as a function of the distance from its centre is represented graphically by -

(1)  (2)

(3)  (4)

Subtopic:  Electric Field |
 67%
AIIMS - 2004
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Suppose the charge of a proton and an electron differ slightly. One of them is \(\text- e\) and the other is \((e+\Delta e)\). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance \(d\) (much greater than atomic size) apart is zero, then \(\Delta e\) is of the order: [Given the mass of hydrogen, \(m_h = 1.67\times 10^{-27}~\text{kg}\)]
1. \(10^{-20}~\text{C}\)
2. \(10^{-23}~\text{C}\)
3. \(10^{-37}~\text{C}\)
4. \(10^{-47}~\text{C}\)

Subtopic:  Coulomb's Law |
 63%
From NCERT
NEET - 2017
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Two identical charged spheres suspended from a common point by two massless strings of lengths l are initially at a distance d(d < < l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then, v varies as a function of the distance x between the sphere, as:

(a) \(v \propto x\)

(b) \(v \propto x^{\frac{-1}{2}}\)

(c) \(v \propto x^{-1}\)

(d) \(v \propto x^{\frac{1}{2}}\)
 

Subtopic:  Coulomb's Law |
 68%
From NCERT
NEET - 2016
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The electric field in a certain region is acting radially outward and is given by E=Ar. A charge contained in a sphere of radius 'a' centered at the origin of the field will be given by

1. 4πoAa2

2. πoAa2

3. 4πoAa3

4. oAa2

Subtopic:  Gauss's Law |
 70%
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. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become:

1. (1/√2)2

2. (r/23) 

3. (2r/√3)

4. (2r/3)

Subtopic:  Coulomb's Law |
 76%
From NCERT
NEET - 2013
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An electric dipole of moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle θ with the direction of the field. Assuming that the potential energy of the dipole to be zero when θ=90°, the torque and the potential energy of the dipole will respectively be

1. pE sinθ,-pE cosθ

2. pE sinθ,-2pE cosθ

3. pE sinθ,2pE cosθ

4. pE cosθ,-pE sinθ

Subtopic:  Electric Dipole |
 79%
NEET - 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 |
 84%
From NCERT
NEET - 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 being the charge on an electron)

1. 4πεFd2 e2                  2. 4πεFe2d2

3. 4πεFd2e2               4. 4πεFd2q2

 

Subtopic:  Coulomb's Law |
 76%
From NCERT
NEET - 2010
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