(VI) In the description of charge distribution, we often encounter so called multipole expansion, which e.g., describes the electric field, E(r), at distances r. If the system has charge q, then for large distances, i.e., r->infty , the field becomes (q)/(r^(2)). This coincides with the field of a monopole (i.e., a point charge). Yet, if the charge distribution inside the object (e.g., electrons in a molecule) is not homogeneous, we use also dipole, quadrupole, ... and higher order terms, which decay faster than the monopole term but are important when considering how objects like molecules interact. Note that even for a charge neutral system, the dipole and other terms may be nonzero (as the charge distribution may be inhomogeneous).
Investigate the distance dependence of the dipole term illustrated in Figure 1 below for a trivial "1D problem". The dipole of a system is defined as distribution of N point charges with respect to their center:
p=sum_(j=1)^N q_(j)x_(j).
The total electric field along the horizontal axis in this case is:
E(r)=sum_(j=1)^2 (q_(j))/(4pi epsi lon_(0))((1)/(r_(j)^(2)))
where epsi lon_(0) is the vaccum permittivity constant and r_(j) is the distance to charge q_(j).
First, show that in the case illustrated in Figure 1:
E(r)=(q)/(4pi epsi lon_(0)r^(2))((1)/((1+(x)/(r))^(2))-(1)/((1-(x)/(r))^(2)))
Now, consider the two charges as a single object and take the first-order binomial Taylor expansion of this electric field expression and truncate it at the linear order in x (which is valid for rā«x ); next, express the dependence of the electric field on the distance r and the dipole p. Your expression should look like this:
E(r)=(q)/(4pi epsi lon_(0))(2p)/(r^(m))
determine the power m in the denominator.
Finally: Neutral molecules can be held together by weak intermolecular (Van der Waals) forces, which can be thought of as caused by dipole-dipole interactions (i.e., prop|E(r)^(2)| induced due to charge fluctuations. Argue, why these interactions need to decay as fast as r^(-6).
(VI In the description of charge distribution we often encounter so called multipole expan- sion,which c.g.,describes the electric field,Er), at distances r. If the system has charge q,then for large distances,i.e.r oo, the ficld becomes q/r2.This coincides with the field of a monopole (i.e., a point charge. Yet, if the charge distribution inside the object (e.g., electrons in a molecule is not homogeneous, we usc also dipole, quadrupole,... and higher order terms, which decay faster than the monopole term but are important when considering how objects like molecules interact. Note that even for a charge neutral system, the dipole and other terms may be nonzero (as the charge distribution may be inhomogeneous). Investigate the distance dependence of the dipole term illustrated in Figure 1 below for a trivial "1D problem". The dipole of a system is defined as distribution of N point charges with respect to their center:
qjEj
The total electric field along the horizontal axis in this case is:
E(r=Dq j=1 4TTEO
where co is the vaccum permittivity constant and r, is the distance to charge qj First,show that in the case illustrated in Figure 1:
b - Er= 4TEOT2 1+2
1
Now,consider the two charges as a single object and take the first-order binomial Taylor expansion of this clectric field expression and truncate it at the linear order in x (which is valid for r>);next,express the dependence of the electric field on the distance r and the dipole p. Your expression should look like this:
q2p
determine the power m in the denominator. Finally: Neutral molecules can be held together by weak intermolecular (Van der Waals) forces,which can be thought of as caused by dipole-dipole interactions i.c. |Er induced due to charge fluctuations. Argue, why these interactions need to decay as fast as T-6