Potential Energy of a Point Charge Placed at a Point with Non Zero Potential
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- 24 cm from+9e
- 12 cm from+9e
- 24 cm from+e
- 12 cm from+e
When an alpha particle of mass M moving with velocity V bombards a heavy nucleus of charge Ze its distance of closest approach from the nucleus depends on M As:
- 6 cm from +9q
- 12 cm from +9q
- 6 cm from +q
- 12 cm from +q
- 1 MeV
- 8 MeV
- 4 MeV
- 2 MeV
- 1.6×10−24 joule
- 1.6×10−14 joule
- 0.53×10−14 joule
- 1.6×10−14 joule
- 3.2×10−10J
- 3.2×10−18J
- 1J
- 1 dyne
- −0.6 mJ
- 0.4 mJ
- 4 μJ
- 6 μJ
- 0.37 m from 4 nC charge toward left
- 0.47 m from 4 nC charge toward left
- 0.57 m from 4 nC charge toward left
- 0.67 m from 4 nC charge toward left
An electron (mass) is sent into an electric field of intensity . The acceleration produced is
The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude, the distance being measured from its equilibrium position is
2 : 1
3 : 1
8 : 1
4 : 1
- 1000 ergs
- 1000 joules
- 1000 kWh
- 500 ergs
- 2.5×107 Vm−1
- 2.5×10−7 Vm−1
- 5×10−7 Vm−1
- 5×107 Vm−1
- Zero
- q(Q1−Q2)(√2−1)√2(4πε0R)
- q(Q1−Q2)(√2+1)√2(4πε0R)
- q(Q1+Q2)(√2−1)√2(4πε0R)
- 3q2a24πε0d3
- 3q2a2πε0d3
- q2a27πε0d3
- q2a22πε0d3
Electric potential in electrostatic is analogous to _________ in heat.
- 16 V
- 8 V
- −8 V
- −16 V
- Q22KAεo
- Q2KAεo
- 2Q2KAεo
- Q24KAεo
- 180 V
- 360 V
- 90 V
- 18 V
- 2.5×107 Vm−1
- 2.5×10−7 Vm−1
- 5×10−7 Vm−1
- 2.5×10−7 Vm−1
- 99J
- 90J
- 100J
- 1J
The blue charge is negative. The red charge is positive.
Answer are expressed in terms of k (the coulomb constant) and given distances
- k2d
- −k2d
- k4d
- −kd
- −k4d
(a) Charge −q is distributed on the surfaces as
(A) −Q on the inner surface, −Q on outer surface
(B) −Q on the inner surface, −q+Q on outer surface
(C) +Q on the inner surface, −q−Q on outer surface
(D) The charge −q is spread uniformly between the inner and outer surface.
(b) Assume that the electrostatic potential is zero at an infinite distance from the spherical shell. The electrostatic potential at a distance R(a<R<b) from the centre of the shell is [whereK=14πε0]
- 0
- KQa
- KQ−qR
- KQ−qb
Two equal charges q of opposite sign are separated by a small distance '2l'. The electric potential at a point on the perpendicular bisector of the line joining the two charges at a distance r is :
- Zero
- 14πε02qr
- 14πε0qr
- 14πε02qr2
- zero
- q(Q1−Q2)(√2−1)4√2πϵ0r
- q(Q1−Q2)(√2+1)4√2πϵ0r
- q√2(Q1+Q2)4√2πϵ0r
Equal and similar charges are placed on the earth and the moon. The numerical value of charge required to neutralise their gravitational attraction is (Given: mass of earth=6×1024kg, Mass of moon=7×1022kg) :
- 15.3×10−13C
- 9.3×1013C
- 5.5×1013C
- 3.6×1013C
The kinetic energy of the particle after time ‘t’ is
- 2E2t2mg
- E2mq22t2
- Eqm2t
- E2q2t22m
If a 3C charge moves from a point of potential V1 to another point of potential V2, the change in P.E of the system will be?
3V1
3(V2−V1)
- 3V2
3(V1−V2)
Where is the electric potential zero?