Home BSc Physics Moment Of Charge Distribution , Multipole Expansion
Moment Of Charge Distribution , Multipole Expansion
Let us consider a charge in an irregular system. For determination of potential at an outer point, we assume the entire system is made up of small volume elements.
\[
\rho=\frac{dq}{dV}
\]
Here, dq = charge in small volume element dV
Let, Distance of volume element from origin = r , Distance of point P from origin = R , Distance between volume element dV and point P = a
Potential Due to Small Volume Element
We know that potential at point P due to dV is:
\[
d\phi=
\frac{k\rho dV}{a}
\qquad -------(1)
\]
Integrating both sides for total potential
\[
\int d\phi=
\int
\left(
\frac{k\rho dV}{a}
\right)
\]
Therefore,
\[
\phi=
k
\int
\frac{\rho dV}{a}
\qquad -----(2)
\]
Finding Distance a
Suppose volume element makes angle θ with z-axis. In triangle OP(dV),
OP = R , OdV = r , PdV = a
Applying cosine rule:
\[
a^2=
R^2+r^2-2Rr\cos\theta
\]
Therefore,
\[
\frac1a=
\frac1{
\sqrt{
R^2+r^2-2Rr\cos\theta
}}
\]
Taking R² as common:
\[
\frac1a=
\frac1R
\left[
1+
\frac{r^2}{R^2}
-
\frac{2r\cos\theta}{R}
\right]^{-1/2}
\]
Using Binomial Expansion
We know:
\[
(1+x)^n=
1+nx+
\frac{n(n-1)x^2}{2!}
+
\frac{
n(n-1)(n-2)x^3
}{3!}
+...........
\]
Now,
\[
\frac1a=
\frac1R
\left[
1-
\frac12
\left(
\frac{r^2}{R^2}
-
\frac{2r\cos\theta}{R}
\right)
+
\frac38
\left(
\frac{r^2}{R^2}
-
\frac{2r\cos\theta}{R}
\right)^2
+............
\right]
\]
After simplification:
\[
\frac1a=
\frac1R+
\frac{r\cos\theta}{R^2}
+
\frac{r^2}{R^3}
\left[
\frac{3\cos^2\theta-1}{2}
\right]
+...........................(4)
\]
Substituting Equation (4) into Equation (2)
\[
\phi=
k
\left[
\frac1R\int\rho dV+
\frac1{R^2}\int\rho r\cos\theta dV+
\frac1{R^3}
\int
\rho
\frac{(3\cos^2\theta-1)r^2}{2}
dV
+......
\right]
\qquad ...........(5)
\]
Let, consider terms in equitation 5
\[
P_0=\int\rho dV
\]
\[
P_1=
\int\rho r\cos\theta dV
\]
\[
P_2=
\int
\rho
\frac{(3\cos^2\theta-1)r^2}{2}
dV
\]
Therefore,
\[
\phi=
k
\left[
\frac{P_0}{R}
+
\frac{P_1}{R^2}
+
\frac{P_2}{R^3}
+
..........
\right]
\]
We know,
\[
k=
\frac1{4\pi\epsilon_0}
\]
Hence,
\[
\phi=
\frac{P_0}{4\pi\epsilon_0R}
+
\frac{P_1}{4\pi\epsilon_0R^2}
+
\frac{P_2}{4\pi\epsilon_0R^3}
+
................(6)
\]
Physical Meaning of Terms
(i) Monopole Moment
\(P_0=\int\rho dV\)
, This represents total charge present in the system.
Therefore, P₀ = Total Charge If, P₀ = 0 then net charge of system is zero.
(ii) Dipole Moment
\(P_1=\int\rho r\cos\theta dV\)
, This term represents electric dipole moment.
Therefore, P₁ is called Dipole Moment.
(iii) Quadrupole Moment
\(P_2=\int\rho\frac{(3\cos^2\theta-1)r^2}{2}dV\)
, This term is known as Quadrupole Moment.
General Multipole Expansion
Therefore,
\[
\phi=
\frac{P_0}{4\pi\epsilon_0R}
+
\frac{P_1}{4\pi\epsilon_0R^2}
+
\frac{P_2}{4\pi\epsilon_0R^3}
+
\frac{P_3}{4\pi\epsilon_0R^4}
+
..........
\]
This complete series is called "MULTIPOLE EXPANSION" or "MOMENTS OF CHARGE DISTRIBUTION"
Importance of Multipole Expansion
Simplifies complicated charge distributions Useful for far field potential calculations Widely used in: - Electrostatics , Molecular Physics , Nuclear Physics , Antenna Theory , Quantum Mechanics , Gravitational Field Theory
FINAL RESULT:
\[
\phi=
\frac{P_0}{4\pi\epsilon_0R}
+
\frac{P_1}{4\pi\epsilon_0R^2}
+
\frac{P_2}{4\pi\epsilon_0R^3}
+......
\]
This equation represents the MOMENT OF CHARGE DISTRIBUTION or MULTIPOLE EXPENSION
