## Wednesday, June 16, 2021

### Basic Electrostatics - An Introduction (Part 1)

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 eo r2

(Where eo 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 eo r}  =       1 / 4p eo  [Q/ r2 ]   (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)/ x2

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)/ x2

W = -  Q q/ 4p e    [-1/ x2 ]¥ r   = Q q/ 4p eo 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 eo r     (J/C or Volts, V)

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

VBC = Q / 4p e(1/rB – 1/rC) =   Q / 4p e(1/r – 1/¥)

VBC = Q / 4p eo 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/ eo.

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

Fig. 3

Let the radius of the larger sphere be r, and then the surface area A =  4pr2. The flux through the surface is defined: F =  E x (Area) = E x (4pr2).

But E =  Q / 4p eo  r2     x  (  4pr2) = Q / eo.

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 4th (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.