The minimum velocity v with which charge q should be projected so that it manages to reach the centre of the ring starting from the position shown in figure is
The work done in moving a positive test charge qo from infinity to a point P at a distance r from the charge q is
Definition based
Electric field intensity is equal to
Definition based
Electric potential at a point located far away from the charge is taken to be
Electric potential at a point located far away from the charge is taken to be zero Because the distance is much larger so in other words there is no effect of other charges on that charge.
A charge of 6 mC is located at the origin. The work done in taking a small charge of 2 x 10^{9} C from a point P (0, 3 cm, 0) to a Q (0,4 cm, 0) is
q=6x10^{10} c
Q=2x10^{9}c
r_{1}=3x10^{2}m
r_{2}=4x10^{2}m
△v_{Q}=w/q
(1/4πε_{o})2x10^{9}/10^{2}[(1/4)(1/3)]=W/6x10^{3}
9x10^{9}x2x10^{9}x6x10^{3}/12x10^{2}=W
W=9x10^{3}x10^{2}
W=0.9J
A particle of mass 1 Kg and charge 1/3 μC is projected towards a non conducting fixed spherical shell having the same charge uniformly distributed on its surface. The minimum intial velocity V_{0} of projection of particle required if the particle just grazes the shell is
From conservation of angular momentum,
mr_{0} = r/2 = mvr
⇒ v = v_{0}/2
From conservation of energy,
On moving a charge of 20 coulombs by 2 cm, 2 J of work is done, then the potential difference between the points is
Potential difference between two points is given by
V_{a}  V_{b} = W/q_{0}
Work, W = 2 J
Charge, q_{0} = 20 C
Potential difference = 2/20 = 0.1 V
The correct option is C.
The amount of work done in moving a charge from one point to another along an equipotential line or surface charge is
Since Potential difference between two points in equipotential surfaces is zero, the work done between two points in equipotential surface is also zero.
A charge is uniformly distributed inside a spherical body of radius r_{1} = 2r_{0} having a concentric cavity of radius r_{2} = r_{0} (ρ is charge density inside the sphere). The potential of a point P or a distance 3r_{0}/2 from the centre is
Electric field at a distance r,
(r_{0} < r < 2r_{0}) from the center is given by
Potential at the outer surface,
∴ If V is the required potential at r = 3r_{0}/2,
Electric field intensity at point ‘B’ due to a point charge ‘Q’ kept at a point ‘A’ is 12 NC^{1} and the electric potential at a point ‘B’ due to same charge is 6 JC^{1}. The distance between AB is
E.l = V where,
E = electric field intensity = 12 N/C
V = electric potential = 6 J/C
=> distance between A and B,
l = (6 / 12) m or (1 / 2) m = 0.5 m
E = dV/dr, here the negative sign signifies that
Answer : d
Solution : The negative sign is just a convention and it signifies that the direction of E is opposite to the direction in which potential increases.
Work done in carrying 2C charge in a circular path of radius 2m around a charge of 10C is
The overall work performed in carrying a 2coulomb charge in a circular orbit of radius 3 m around a charge of 10 coulomb is calculated below.
It is a wellknown fact that W=qdv.
Here dV is the change in overall potential. In the circular orbit of r potential at each point is similar.
Most significantly, the value of r is 3.
The value of dv=0 and hence W=q_{0}=0.
Dimensional formula for potential difference is
A point charge 2nC is located at origin. What is the potential at (1,0,0)?
V = Q/(4πεr), where r = 1m
V = (2 X 10^{9})/(4πε x 1) = 18 volts.
If 100 J of work has to be done in moving an electric charge of 4C from a place where potential is 5 V to another place, where potential is V volt. The value of V is
From the definition, the work done to a test charge ‘q0’ from one place to another place in an electric field is given by the formula
W=q_{0}x[v_{final}v_{initial} ]
100=4x[v(5)]
v+5=25
v=20V
A particle of charge q_{1} = 3μC is located on xaxis at the point x_{1} = 6 cm. A second point charge q_{2} = 2μC is placed on the xaxis at x_{2} = 4 cm. The absolute electric potential at the origin is
V=kq/r
Therefore, V_{net}=Kx[(2µc/4) + (3µc/6)]
=9c10^{9}x10^{6}[1/2+1/2]x10^{2}
=9x10^{5}
Electric potential is
Electric potential is the electric potential energy per unit charge. In equation form, V=U/q, where U is the potential energy, q is the charge, and V is the electric potential. Since both the potential energy and charge are scalar quantities, so does the potential.
And it's dimension [ L^{2}M^{1}T^{3}I^{1}]
A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral.
When the charge is given to inner cylinder than there is an electric field is produced between cylinders which is given by
and due to this a potential difference is developed between two cylinders.
Equal charges are given to two spheres of different radii. The potential will
When equal charges are given to two spheres of different radii, the potential will be more or the smaller sphere as per the equation, Potential = Charge / Radius.
Since potential is inversely proportional to radius, the smaller radius will have higher potential and vice versa.
Consider a solid cube made up of insulating material having a uniform volume charge density. Assuming the electrostatic potential to be zero at infinity, the ratio of the potential at a corner of the cube to that at the centre will be
By dimensional analysis
but by superposition
Because of the centre of the larger cube lies at a corner of the eight smaller cubes of which it is made
therefore,
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