An elementary particle of mass m and charge e is projected with velocity v at a much more massive particle of charge Ze, where . What is the closest possible approach of the incident particle ?

(1) 

(2) 

(3) 

(4) 

Subtopic:  Electric Potential Energy |
 57%
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Four equal charges Q are placed at the four corners of a square of each side is ‘a’. Work done in removing a charge – Q from its centre to infinity is 

(1) 0

(2) 2Q24πε0a

(3) 2Q2πε0a

(4) Q22πε0a

Subtopic:  Electric Potential Energy |
 52%
From NCERT
AIIMS - 1995
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Two spheres of radius a and b respectively are charged and joined by a wire. The ratio of the electric field at the surface of the spheres is 

(1) a/b

(2) b/a

(3) a2/b2

(4) b2/a2

Subtopic:  Electric Potential |
 55%
From NCERT
PMT - 1999
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An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. The final speed of the electron will be 

(1) Ve/m

(2) eV/m

(3) 2eV/m

(4) 2eV/m

Subtopic:  Electric Potential Energy |
 83%
From NCERT
PMT - 2000
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The dimension of (1/2) ε0E2  (ε0: permittivity of free space; E: electric field) is

(1) MLT–1

(2) ML2L–2

(3) ML–1T–2

(4) ML2T–1

Subtopic:  Energy stored in Capacitor |
 74%
From NCERT
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A table tennis ball that has been covered with conducting paint is suspended by a silk thread so that it hangs between two plates, out of which one is earthed and other is connected to a high voltage generator. This ball

(1) Is attracted towards high voltage plate and stays there

(2) Hangs without moving

(3) Swing backward and forward hitting each plate in turn

(4) Is attracted to the earthed plate and stays there

Subtopic:  Capacitance |
From NCERT
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Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle θ with the axis of the dipole, then the potential at P is given by (r >> 2a) (Where p = 2qa

(1) V=pcosθ4πε0r2

(2) V=pcosθ4πε0r

(3) V=psinθ4πε0r

(4) V=pcosθ2πε0r2 

Subtopic:  Electric Potential |
 71%
From NCERT
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A charge \(+q\) is fixed at each of the points x=x0, x=3x0, x=5x0 ..... infinite, on the \(x\)-axis, and a charge \(-q\) is fixed at each of the points x=2x0, x=4x0,x=6x0,..... infinite. Here \(x_0\) is a positive constant. Take the electric potential at a point due to a charge \(Q\) at a distance \(r\) from it to be \(\frac{Q}{4\pi \varepsilon_0 r}\). Then, the potential at the origin due to the above system of charges is:
1. \(0\)
2. \(\frac{q}{8 \pi \varepsilon_{0} x_{0} \mathrm{ln} 2}\)
3. \(\infty\)
4. \(\frac{q \mathrm{ln} 2}{4 \pi \varepsilon_{0} x_{0}}\)

Subtopic:  Electric Potential |
From NCERT
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Point charge q moves from point P to point S along the path PQRS (figure shown) in a uniform electric field E pointing co-parallel to the positive direction of the x-axis. The coordinates of the points P, Q, R, and S are (a,b,0),(2a,0,0),(a,b,0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression 

(1) qEa

(2) – qEa

(3) qEa2

(4) qE[(2a)2+b2]

Subtopic:  Energy of Dipole in an External Field |
From NCERT
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The electric potential at a point at distance 3R from the centre of disc
of radius R lying in the axis of the disc whose surface charge density is σ
will be given by:
1. σ2ε02-3R                          2. σ2ε02+3R
3.σ2ε03-2R                           4. σ2ε03+2R

Subtopic:  Electric Potential |
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