Electrostatics is an area of physics not covered until now. However, it contains an abundance of principles as well as applications which can be used in many other subdisciplines from electrodynamics and plasma physics to solar physics.

**1.
****Electric Field Lines and Strength:**

** **In
the most basic examination of the electrostatic field, we specify a region
in which a test charge q experiences a force. The most general description
makes use of a spherical region and within it we can posit either a positive
test charge (+ q) or a negative test charge (- q). For the first case, we may
use the field line directions as in Fig.1

*Fig 1.
Field line directions for positive test charge*

**Note that the test charge itself always has positive charge. ** Hence, a positive test charge introduced into
a spherical region with central (+) charge, will always experience a repulsive force away from the central charge,
and along what we deem the E-field lines. For the negative charge region the vector force field lines will be reversed.

And in this case,
the test charge always moves *toward* the central (-) charge, because unlike
charges attract. By *Coulomb’s law*, the force of electrostatic attraction will
be:

F = Q q/ 4p e_{o} r^{2}

** **

(Where
e_{o} is the **permittivity of free space **(8.85 x 10^{-12}
F/m)

** **

**The
electric field strength E is: E = F/**q

Or: **E**
= 1/q { Q q/ 4p e_{o }r^{2
}} = 1 / 4p
e_{o} [Q/ r^{2 }] (N/C)

As seen here, the
electric field strength is the Coulomb force per unit test charge. The field
lines as shown give the direction of the vector ** E** at the
particular point. That is, the direction in which a (+) test charge would
accelerate. While the field lines are imaginary, the field they represent is
very real!

2. ** Electrical
Potential**.

Consider now the
unit test charge +1 to be *on the surface* of the spherical region shown
in Fig.1, for which a positive charge (+) resides at the center. We call this
position ‘A’ and we wish to find the potential here, by reference to Fig. 2
below.

**.**

*Fig. 2: Situation to find Electrical potential at point A*Note that in
traversing from C®
B the
direction is opposite to the field E direction so that an external agent must
do work (W = F (-d
x) ) against the force F. As a result the electrical potential must be equal to
the work done.

Note
further that point A is at the same radius as B, and we say the surface is an *equipotential
surface* for which the electrical potential is the same at points B and A.
Thus, the only work done in bringing the test charge in from point C to A is
the same as bringing it from C to B, or:

dW = [Q q/ 4p e ] (-d x)/ x^{2}

^{}

** ****Meanwhile,
the total work done in bringing q from (effectively) infinity to A is;**

ò F (-d x) = - Q q/ 4p
e ò_{¥}_{ }^{r} (d x)/ x^{2}

** **

**W
= - **Q q/ 4p e ** [-1/ x ^{2} ]**

_{¥}**Q q/ 4p e**

^{r}=_{o }r

The potential at A then is the work done per unite positive charge brought from ¥ (at C) to A, Thus:

V = W/ q = Q / 4p e_{o }r (J/C or Volts, V)

With
reference to Fig. 2, the **difference** in potential (V_{B} – V_{C})
is said to be the potential of B with respect to C, abbreviated V_{BC},
or the potential difference for these two points. If C is effectively at ¥, then:

V_{BC} = Q / 4p e_{o }(1/r_{B}
– 1/r_{C}) = Q / 4p e_{o }(1/r
– 1/¥)** **

V_{BC} = Q
/ 4p e_{o }r
= V

Thus, the potential difference is numerically equal to the potential difference between the point and infinity, which implies both the potential and potential difference are measured in volts.

3. **Gauss’ Law**

**One
of the more important theorems in electrostatics is Gauss’ law: The flux through a spherically symmetric
region bearing a charge +Q is equal to the charge divided by the permittivity
of free space, or Q/ **e_{o.}

**Proof: **We refer to the diagram in Fig. 3:

Let
the radius of the larger sphere be r, and then the surface area A = 4pr^{2}. The** flux** through the surface is
defined: F = E x (Area) = E x (4pr^{2}).

But
E = Q / 4p e_{o} r^{2
} x ( 4pr^{2}) = **Q / **

*e*

_{o}_{.}

**Thus,
the total flux crossing any surface drawn outside and concentrically around a
point charge is a constant. (But only if the inverse square law, i.e. Coulomb’s
law, is valid). **

**Suggested Problems:**

**1)
****A particle of charge +3 x 10 ^{-9}
C is in a uniform E-field directed to the left:**

**¬****E ----------------------x q--------**

It is released from rest and
moves a distance of 5 cm, after which its kinetic energy is found to be 4.5 x
10^{-5} J.

(a)What is the magnitude of the electric field?

(b)What is the potential of the starting point relative to the
endpoint?

2)
A charge of (-20 mmC) is placed at one corner of a 3 x 4 cm rectangle while charges of 10 mmC are placed at two adjacent
corners as shown:

**Using this
data, calculate the potential at the 4 ^{th} (lower left) corner. (Note: potential
depends only on distances, not directions!)**

3) A hollow sphere has a charge of +Q at its center. Find the electric flux from its surface and the charge density**.**

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