Thursday, 31 March 2016

class 12 Notes for Electrostatics (Part 3)

Electrostatic approximation[edit]

The validity of the electrostatic approximation rests on the assumption that the electric field is irrotational:
\vec{\nabla}\times\vec{E} = 0.
From Faraday's law, this assumption implies the absence or near-absence of time-varying magnetic fields:
{\partial\vec{B}\over\partial t} = 0.
In other words, electrostatics does not require the absence of magnetic fields or electric currents. Rather, if magnetic fields or electric currentsdo exist, they must not change with time, or in the worst-case, they must change with time only very slowly. In some problems, both electrostatics and magnetostatics may be required for accurate predictions, but the coupling between the two can still be ignored. Electrostatics and magnetostatics can both be seen as Galinean limits for electromagnetism.[2]

Electrostatic potential[edit]

Because the electric field is irrotational, it is possible to express the electric field as the gradient of a scalar function,\phi, called the electrostatic potential (also known as the voltage). An electric field, E, points from regions of high electric potential to regions of low electric potential, expressed mathematically as
\vec{E} = -\vec{\nabla}\phi.
The Gradient Theorem can be used to establish that the electrostatic potential is the amount of work per unit charge required to move a charge from point a to point b is the following line integral:
-\int_a^b{\vec{E}\cdot \mathrm{d}\vec \ell} = \phi (\vec b) -\phi(\vec a).
From these equations, we see that the electric potential is constant in any region for which the electric field vanishes (such as occurs inside a conducting object).

Electrostatic energy[edit]

Main articles: Electric potential energy and Energy density
A single test particle's potential energy,  U_\mathrm{E}^{\text{single}}, can be calculated from a line integral of the work, q_n\vec E\cdot\mathrm d\vec\ell . We integrate from a point at infinity, and assume a collection of N particles of charge Q_n, are already situated at the points \vec r_i. This potential energy (in Joules) is:
U_\mathrm{E}^{\text{single}}=q\phi(\vec r)=\frac{q }{4\pi \varepsilon_0}\sum_{i=1}^N \frac{Q_i}{\left \|\mathcal{\vec R_i} \right \|}
where  \vec\mathcal {R_i} = \vec r - \vec r_i is the distance of each charge Q_i from the test charge q, which situated at the point \vec r , and  \phi(\vec r) is the electric potential that would be at \vec r  if the test charge were not present. If only two charges are present, the potential energy is k_eQ_1Q_2/r. The total electric potential energy due a collection of N charges is calculating by assembling these particles one at a time:
U_\mathrm{E}^{\text{total}} = \frac{1 }{4\pi \varepsilon _0}\sum_{j=1}^N Q_j \sum_{i=1}^{j-1} \frac{Q_i}{r_{ij}}= \frac{1}{2}\sum_{i=1}^N Q_i\phi_i ,
where the following sum from, j = 1 to N, excludes i = j:
\phi_i = \frac{1}{4\pi \varepsilon _0}\sum_{j=1 (j\ne i)}^N \frac{Q_j}{r_{ij}}.
This electric potential, \phi_i is what would be measured at \vec r_i if the charge Q_i were missing. This formula obviously excludes the (infinite) energy that would be required to assemble each point charge from a disperse cloud of charge. The sum over charges can be converted into an integral over charge density using the prescription \sum (\cdots) \rightarrow \int(\cdots)\rho\mathrm d^3r:
U_\mathrm{E}^{\text{total}} = \frac{1}{2} \int\rho(\vec{r})\phi(\vec{r}) \operatorname{d}^3 r = \frac{\varepsilon_0 }{2} \int \left|{\mathbf{E}}\right|^2 \operatorname{d}^3 r,
This second expression for electrostatic energy uses the fact that the electric field is the negative gradient of the electric potential, as well asvector calculus identities in a way that resembles integration by parts. These two integrals for electric field energy seem to indicate two mutually exclusive formulas for electrostatic energy density, namely \frac{1}{2}\rho\phi and \frac{\varepsilon_0 }{2}E^2; they yield equal values for the total electrostatic energy only if both are integrated over all space.

Electrostatic pressure[edit]

On a conductor, a surface charge will experience a force in the presence of an electric field. This force is the average of the discontinuous electric field at the surface charge. This average in terms of the field just outside the surface amounts to:
P = \frac{ \varepsilon_0 }{2} E^2 ,
This pressure tends to draw the conductor into the field, regardless of the sign of the surface charge.
rest continue in part 4.
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