Contents
Physics Topics can also be used to explain the behavior of complex systems, such as the stock market or the dynamics of traffic flow.
What do you mean by Potential of a Charged Body and Potential due to Electric Dipole?
i) In CGS system: The CGS unit of potential is esu of poten-tial or statvolt (statV).
The potential at a point in an electric field is said to be 1 esu if 1 erg of work is done in bringing 1 esu of positive charge from infinity to that point.
ii) In SI: The SI unit of potential is volt (V).
The potential at a point in an electric field is said to be 1 volt if 1 joule of work is done in bringing 1 coulomb of positive charge from infinity to that point.
1V = \(\frac{1 \mathrm{~J}}{1 \mathrm{C}}\) = \(\frac{10^7 \text { erg }}{3 \times 10^9 \text { esu of charge }}\)
= \(\frac{1}{300} \times \frac{1 \text { erg }}{1 \text { esu of charge }}\)
= \(\frac{1}{300}\) esu of potential
∴ 1 esu of potential = 300 volt
The unit of potential difference is the same as that of potential.
Dimension of Potential:
Potential of the Earth
The potential of a body is measured with reference to the potential of the earth which is taken as the standard and conventionally assigned the value of zero. The earth is so Large
that its potential does no change appreciably due to small gain or loss of charge. It effectively maintains its constant potential. A body is said to be at a positive potential if its potential is above that of the earth and at a negative potential if its potential is below that of the earth. This is analogous to the situation that sea level is taken as the standard zero level to measure the altitudes and depths of different places on the earth.
Potential of a Charged Body
A body is said to be at a positive potential if there is a flow of electrons from the earth to it when the body comes in contact with the earth [Fig.] If electrons flow from the body to the
earth when an electrical contact is established between them, the body is said to be at a negative potential. In both the cases the flow of electrons will continue until the potential of the body becomes zero, i.e., equal to the potential of the earth. Any earthed (or grounded) conductor is effectively at zero potential.
Measure of potential of a charged body: The work done by an external force to bring a unit positive charge without acceleration very near to a charged body from infinity is the measure of potential of that body.
Potential at a Point in the Field of a Point Charge
Let us consider a charge +q at a point A in vacuum or air.
Let P be a point at a distance r from A, where the electric potential due to the charge at A is to be determined.
The intensity at P, in vacuum or air, due to the charge +q at A = \(\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}\) ; along \(\overrightarrow{A P}\) [Fig.]
If PP1 = dr be a very small distance beyond P where the intensity effectively remains che same, then work done by the external force (equal hut opposite of the electric force) in bringing a unit positive charge from P1 to P is,
dW = external force acting on the unit positive charge × its displacement
= \(\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \cdot(-d r)\) [The negative sign is taken because intensity and displacement are oppositely directed]
Therefore, the work done in bringing the unit positive charge from infinity to P is given by,
So, potential at P due to the charge +q at A is,
V = \(\frac{1}{4 \pi \epsilon_0} \frac{q}{r}\) …. (1)
Potential of a point at a distance r from the charge q in a medium of permittivity E is,
V = \(\frac{1}{4 \pi \epsilon} \frac{q}{r}\) = \(\frac{1}{4 \pi \kappa \epsilon_0} \frac{q}{r}\)
In CGS system, for vacuum or in air, V = \(\frac{q}{r}\)
Potential difference between two points:
Points are coimear with the charge: Suppose, B and A are two points at a distance r2 and r1 respectively from a point charge + q at O. The points O, B and A lie on the same straight line [Fig.].
According to equation (1), potential difference between B and A is,
It may be noted that potentials of the points equidistant from a point charge are equal. For this reason, potential difference between a point situated on a sphere of radius r2 and a point situated on a sphere of radius r1 may be obtained from equation (2).
ii) For non-collinear points: Let the position vector of a point charge q be \(\vec{r}\)
According to the Fig., if \(\vec{r}_2\) and \(\vec{r}_1\) be the position vectors of B and A respectively then equation (2) can be expressed as,
Potential due to a system of charges: Consider the point charges q1, q2, q3,…… situated at distances r1, r2, r3, …….. from a point P. Since electric potential is a scalar quantity, the algebraic sum of the potentials at P due to individual charges is the potential at P.
If the dielectric constant of a -medium is k, then potential at P due to multiple charges,
V = \(\frac{1}{4 \pi \kappa \epsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\cdots+\frac{q_n}{r_n}\right)\) = \(\frac{1}{4 \pi \kappa \epsilon_0} \sum \frac{q}{r}\)
Potential Due to Electric Dipole
i) Potential at a point on the axis of a dipole (end-on position) : Let AB be an electric dipole formed by the charges +q and -q separated by a small distance 2l [Fig.]. Obviously AB = 21 and dipole moment p = 2lq. The dipole is placed in vacuum or in air. For this dipole, potential at the point P situated on the dipole axis at a distance r from the mid-point O of the dipole is to be determined.
The distance of P from +q charge = (r – l) and that from —q charge = (r + l).
Therefore, potential at P due to the charge +q at B of the dipole is
V1 = \(\frac{1}{4 \pi \epsilon_0} \frac{q}{(r-l)}\) (in SI)
Potential at P due to the charge -q at A of the dipole is
V2 = \(\frac{1}{4 \pi \epsilon_0} \frac{-q}{(r+l)}\) (in SI)
Electric potential is a scalar quantity. Sa the resultant potential at P is
If r \(\gg\) l, l2 can be neglected in comparison to r2.
Then the electric potential at P due to the electric dipole is
V = \(\frac{1}{4 \pi \epsilon_0} \frac{p}{r^2}\) (In SI) ………. (2)
In CGS system equations (1) and (2) are respectively given by,
V = \(\frac{p}{r^2-l^2}\) …. (3)
V = \(\frac{p}{r^2}\) …. (4)
ii) Potential at a point on the perpendicular bisector of a dipole (broadside-on position):
Let AB be an electhc dipole formed by the charges +q and -q separated by a small distance 21 [Fig.]. The dipole is situated in vacuum or in air. Obviously AB = 2l and dipole moment
p = 2lq. For this dipole, potential at P situated on the perpendicular bisector of the dipole at a distance r from the mid-point O of the dipole is to be determined.
Potential at P due to the charge +q at B of the dipole is
V1 = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{B P}\)
Potential at P due to the charge -q at A of the dipole is
V2 = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{-q}{A P}\)
Therefore, the net potential at P is
V = V1 + V2
= \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{B P}+\frac{1}{4 \pi \epsilon_0} \cdot \frac{-q}{A P}\)
= \(\frac{q}{4 \pi \epsilon_0}\left(\frac{1}{B P}-\frac{1}{A P}\right)\) = 0 [∵ BP = AP] … (5)
So the electric potential is zero everywhere on the equatoria) line of an electric dipole (but electric intensity is not zero). Hence this perpendicular bisector is an equipotential
line. No work is done to move a charge along this line.
In CGS system equation (5) remains the same, i.e., V = O.
iii) Potential at any point due to an electric dipole:
Let AB be an electric dipole formed by the charges +q and -q separated by a small distance 2l [Fig.]. Obviously AB = 2l and dipole moment p = 2lq. Let P be a point at a distance r from the mid-point O of the dipole and the line OP make an angle θ with the axis of the dipole i.e., the polar coordinates of P are (r, θ). Potential at P due to the dipole is to be calculated.
Now join PA and PB. Then draw a perpendicular from A which meets the extended OP at D. Also draw BC perpendicular to OP.
If r \(\gg\) l, we can write BP ≈ CP = OP – OC = r – lcosθ
and AP ≈ DP = OP + OD = r + lcosθ
Now, potential at P due to the charge +q at B of the dipole is
Special Case:
i) P is on the axis of the dipole, then θ = 0
or, cosθ = 1
From equation (6) we get, V = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{r^2}\)
This is equation (2).
ii) If P is on the perpendicular bisector, then
θ = 90°
or, cos90° = 0
So from equation (6) we have, V = 0.
This is equation (5).
The vector form of equation (6) is
V = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{\vec{p} \cdot \hat{r}}{r^2}\) ….. (7)
where \(\hat{r}\) = unit vector along \(\vec{r}\).
In CGS system equations (6) and (7) are respectively given by,
V = \(\frac{p \cos \theta}{r^2}\) …… (8)
V = \(\frac{\vec{p} \cdot \hat{r}}{r^2}\) …. (9)