Introduction to Electric Potential:
If a body 'A' applies a force on the other body 'B' and work is done, a transfer of energy takes place that is measured by the work done. For illustration, if we lift a mass, 'm' via a vertical height, 'h', the work done, 'W' by the force we apply is W = mgh. The energy transfer is mgh and we consider that the system gains and store that quantity of gravitational potential energy that is obtained from the conversion of chemical energy by our muscular activity. If the mass falls, the system loses gravitational potential energy, and neglecting the air resistance, there is a transfer of kinetic energy to the mass equivalent to the work done through gravity.
A body if raised above the surface of earth is stated to get potential energy as it can do work in falling. When it is free to move in the gravitational field it will fall to the position in which its potential energy is zero. Likewise, a charged body in an electric field consists of potential energy, and it will tend to move to such parts of the field where its potential energy is smaller. If a positive charge is repelled by the other positively charged body and moves away, its potential energy reduces. It will be zero if it is completely away from the affect of the charged body that is at infinity. We suppose as the zero of electric potential the potential at an infinite distance from any electric charges.
Definition of Electric Potential:
The electric potential at a point in a field can be stated as the work done per unit charge moving from infinity to that point.
You must note that we are always supposing that the charge doesn't influence the field in any manner.
The choice of zero of potential is purely random and whilst infinity might be a few hundred meters in certain cases, in atomic physics where distances of 10-10 m are comprised, it require merely an extremely small distance away from the charge responsible for the field.
Electric potential is a property of a point in a field and is a scalar as it deals by a quantity of work done or potential energy per unit charge. The symbol for potential is 'V' and the unit a joule per coulomb (JC-1) or volts (V).
All point in a field which has the similar potential can be imagined as lying on a surface, termed as an Equipotential surface. If a charge moves on such a surface no energy change takes place and no work is done. The force due the field should thus act at right angles to the equipotential surface at any point.
Thus, Equipotential surfaces and field lines always interest at right angles.
We can now observe that as electric field can thus be represented pictorially by field lines and by means of equipotential surface (or lines in two dimensional diagrams).
The figure above represents the equipotential surfaces for a point charge. These are the concentric spheres (that is, circles in two dimensions).
When equipotential are drawn in such a way that the change of potential from one to the next is constant and then the spacing will be closer where the field is stronger. This obeys from the fact that in order to carry out a certain amount of work in such areas a shorter distance require be travelled.
Potential due to a point charge:
Let assume a positive test charge +q is placed at point O. We have to determine the electric potential at point P at a distance r from point O.
When we move a positive test charge q' from infinity to point P then change in electric potential energy would be:
UP - U∞ = qq'/4πεor
=> Electric potential at point P is:
VP = (UP - U∞)/q' = q/4πεor
=> Potential V at any point due to random collection of point charges is represented by:
V = (1/4πεo) i=1 Σn (qi/ri)
Here we notice that like electric field potential at any point independent of test charge employed to define it.
For continuous charge distributions summation in above expression will be substituted through the integration
V = 1/4πεo ∫(dq/r)
Here dq is the differential element of charge distribution and 'r' is its distance from the point at which 'V' is to be computed.
Potential due to a system of charges:
Potential at a point due to the system of charges is the sum of potentials due to individual charges.
Let suppose a system of charges q1, q2, q3 ...qn having positive vectors r1, r2, r3.....rn relative to the origin P.
The potential V1 at P due to charge 'q1' is given by:
V1 = (1/4πεo) (q1/r1)
Likewise, the potential at P due to charge 'q2' is:
V1 = (1/4πεo) (q2/r2)
We are familiar that the potential at 'P' is the due to total charge configuration and is the algebraic sum of potentials due to individual charges (By using the superposition principle).
V = V1 + V2 + V3 + ..... + Vn
V = (1/4πεo) [(q1/r1) + (q2/r2) +.......... (qn/rn)]
We know that for a uniformly charged spherical shell, the electric field outside the shell is as if the whole charges are concentrated at the centre. Therefore, the potential outside the shell is represented by:
V = (1/4πεo) (q/r) (r > R)
Here 'q' is the net charge on the shell and 'R' the radius. Electric field within the shell is zero that implies potential is constant inside and equivalent to its value at the surface.
The electric potential difference among the two points is simply the energy needed to transport a unit charge among those two points. This is measured in volts. One volt is stated as one joule per coulomb. It could as well be stated as the change in the potential energy which takes place due to transport of a unit charge from one point to the other. The Electric current flows from a point of high electric potential, to a point that consists of a lower potential. It is just similar to the behavior of water which for all time seeks its level. Current flows among the two points which have a difference in electric potential, to balance the inequality.
As the current flows among two points, each and every charge gains energy which is equivalent to the difference of potential between them. The two terminals of any battery encompass a potential difference between them that is measured in volts. If you connect any circuit between the two terminals of the battery, the charge flows from one terminal to the other, to equalize the charge imbalance and carries on till the difference between them is equalized.
The amount of work done in bringing a unit positive charge from infinity, first to point A and then to point B. Keep in mind that point A and B are in the field of charge 'q'.
Potentials at point A and B are:
VA = q/4πεoϒA
And VB = q/4πεoϒB
The difference of such two potentials (that is, VB - VA) is the work done in taking a unit charge from A to B, and it is termed as the potential difference the two points B and A, it is written as
VBA = VB -VA = q/4πε [(1/ϒB) - (1/ϒA)]
We notice that the work done in carrying the charge in an electric field is independent of the path. It is just this path independence which enables us to state the concept of potential.
If a unit positive charge transports a charge q' between A and B, then the work 'W' done is given by:
W = q'VBA = q' (VB - VA)
Relation between Electric Field and Electric Potential:
Let us consider a charge +Q at a point 'A' in an electric field where the field strength is 'E'. The force, F on Q is given by:
F = EQ
When Q moves a very short distance δx from A to B in the direction of E, then the work done δW by the electric force on q is:
δW = force x distance
δW = Fδx
δW = EQδx (supposing E is constant over AB)
When the potential difference between B and A is δV, then we have by the definition of potential difference:
δV = work done per unit charge
δV = δW/Q = - EQ/Q (δx)
That is δV = -Eδx
The negative sign is inserted to represent that when displacements in the direction of 'E' are taken to be positive, then when δx is positive, δV is negative, that is, the potential reduces.
In the limit, as δx ----01 E becomes the field strength at a point (A). In calculus notation
E = Lim δx---0 (δV/δx) = dv/dx
dv/dx is termed as the potential gradient in the direction and so the field strength at a point equivalents the negative of the potential gradient there.
This is measured in volts per metre (Vm-1). The Vm-1 and NC-1 are both units of E; however the Vm-1 is the one which is generally employed.
If the electric field 'E' is uniform, which is constant in magnitude and direction at all points, it follows that:
dv/dx is constant
that is, dv/dx = - E
∴ V = Ex
∴ E = - v
Electric Field and Potential of an Electric Dipole:
A pair of equivalent and opposite charge, ± q, separated through a vector distance a is termed a dipole. The vector a, which is as well all along the axis of the dipole, is drawn from the negative to the positive charge.
A molecule comprising of a negative and positive ion is an illustration of an electric dipole in nature. An atom comprises of equivalent amount of positive and negative charges whose centre coincides, therefore an atom is neutral for all points outside it. Though, in the presence of an external electric field, the centers of negative and positive charges get separated. The atom then becomes a dipole.
1) Electric Field at a point P all along the axis of the dipole:
E ≈ 2 P r ˆ/4πεor3 for r >> a
Here p = qa rˆ
2) Potential due to a dipole:
Vp = (p. r-)/(4πεor2) = (P cosθ)/(4πεor2)
Vp = 0 when θ = π/2 (P lies on the perpendicular bisector dipole axis)
We can conclude from both the equation that:
a) The dipole potential differs as 1/r2 and the field as 1/r3 as compared to a point charge for which the potential differs as 1/r and the field as 1/r2.
Therefore the potential and field reduce more rapidly having 'r' a dipole than for a point charge.
b) The dipole potential disappears on points which lie on the perpendicular bisector of the dipole axis. Therefore no work is done in moving a test charge all along the perpendicular bisector.
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