Electrostatics

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Coulomb's Law

You've heard that like charges repel while unlike charges attract. There seems to be a force that pulls them together or further apart. What would happen if the charges grew in strength or moved closer together? This could be explained by Coulomb's Law which describes the relationship between the force,F the distance between point charges,r and quantity of charge,q.


File:Illustration.png


F=\frac{kq_{1}q_{2}}{r^{2}}

where

q = quantity of charge (C)

r = separation (m)

k=\frac{1}{4\pi\epsilon _{0}r^{2} }

{\epsilon _{0}=8.85\times 10^{-12}\text{Fm}^{-1}}


Combining these gives 

  F=\frac{q_{1}q_{2}}{4\pi \epsilon _{0}r^{2}} Applicable in vacuum or in air.

This applies in vacuum or free space since ε0 is the permittivity of free space. It is also applicable in air since the permittivity of air is close to that of vacuum.


For other mediums, the formula is modified to include the relative permittivity of the medium, ε 

   F=\frac{q_{1}q_{2}}{4\pi\epsilon \epsilon _{0}r^{2}} Applicable in other insulating mediums.


Example 1

A case of two positive charges in a straight line


Determine the force acting between a 3 μ C charge and a 6 μ C charge separated by 1cm.


3\mu C= 3\times 10^{-6}

6\mu C=6\times 10^{-6}


F=\frac{kq_{1}q_{2}}{r^{2}} =\frac{ 3\times 10^{-6}\times 1 \times 10^{-6}} {4\pi\times 8.85\times 10^{-12}\times \left ( 1\times 10^{-2} \right )^{2}}

F = 2.698 \times 10^{2} \text {N} 

 Notice that the positive value implies a repulsive force


Example 2

Resultant force on a charge due to other charges along a straight line

File:Coulomb2.png

Calculate the resultant force acting on q2 due to both q1 and q3 indicating also its direction.

Both q1 and q3 repels q2 since they are all positive charges. The resultant force is a combination of Fa ( Force on q2 due to q1) and Fb( Force on q2 due to q3)

F_{a}=\frac{k(1\mu \times 2\mu )}{0.001^{2}}


F_{b}=\frac{k(2\mu \times 3\mu )}{0.002^{2}}


The repulsive force due to a charge will push q2 away from that charge.

Sketch the direction of these forces to help determine the direction of the resultant force

File:Coulomb2bx.png


FResultant = Fa − Fb = 4.5kN ( to the right)


Example 3 Resultant force due to charges at angles

Determine the resultant force acting on Qb due to Qa and Qc.


File:CoulombEg3.png

Qa = +3 μC

Qb = +4 μC

Qc = -5 μC


When determining the resultant force, it is best to calculate magnitude of each component forces first and then determine the direction of the resultant force by considering the vectors.

Sketch the forces acting on Qb

File:CoulombEg3ii.png

Determine the magnitude of Fa and Fb.


F_{a}=\frac{k(3\mu \times 4\mu )}{0.0005^{2}}


F_{b}=\frac{k(4\mu \times 5\mu )}{0.001^{2}}


Sketch the direction of the resultant force, Fr and determine its magnitude using Phythagoras theorem.

File:CoulombEg3iii.png

F_{r}=\sqrt{F_{a}^{2}+F_{b}^{2}}


F_{r}= 1.68\times 10^{5} N


The resultant force acts at an angle , X to the line connecting Qb and Qc as shown in the diagram.


Angle X can be found using trigonometry.


X=\tan^{-1}\left ( \frac{F_{b}}{F_{a}} \right )

X=\tan^{-1}\left ( \frac{4.32\times 10^{5}}{1.80\times 10^{5}} \right )

X = 67.40


Electric Field

We know that there is electrical force between charges as expressed by Coulomb's law. The region or space where a charged object would experience that electric force is known as the electric field.

The electric field strength, E is defined as force per unit charge.

E=\frac{F}{q} (NC − 1or Vm − 1)

Physically, higher electric field strength would cause a charge to be subject to greater force than if it were in a field with lower electric field strength.

Electric field strength is a vector. Its direction follows the expected path of a free positive charge. It helps to imagine a single positive charge being dropped near a fixed charge or charges. The path it takes defines the electric field line.


By substituting the electric force, F into the definition of E we come to the formula for E

E=\frac{Q}{4\pi \epsilon _{0}r^{2}}


Example

Electric field strength at a point

Determine the magnitude and direction of the electric field strength at the point P due to the point charges at A and B as shown in the diagram.


File:Efield1.png

Charge A : +1µC , Charge B : -1µC

In the same way as we've sketched the component forces and calculated the resultant force in the section on Coulomb's law, determine the component electric field vectors and its resultant.

Sketch the E vectors starting from point P. You do this by drawing the expected path in reaction to individual source charges as if there is a positive charge at point P.

File:Efield2.png

Calculate the magnitude of component E vectors. Substitute the source charge into Q


E_{1}=\frac{1\times 10^{-6}}{4\pi \varepsilon (10\times 10^{-2})^{2}}

E_{2}=\frac{1\times 10^{-6}}{4\pi \varepsilon (10\times 10^{-2})^{2}}


Rearrange the vectors without changing their angle to form a triangle.

File:Efield3.png

Calculate the resultant E vector by using Phythagoras theorem

E_{r}=\sqrt{E_{1}^{2}+E_{2}^{2}}


Determine the direction of the resultant vector, angle z

Z = tan-1 (E1/E2)

The electric field acting at point P due to charges A and B is 1.27\times 10^{7} N acting in the direction z = 45o as shown in the diagram.


Gauss's Law

Electric flux can be understood as the number of electric field lines from a source going through an area at a 90° angle.

Electric flux, Φ going through a perpendicular area, A is given by ;

Φ = EA

where E : The electric field strength

Unit : Nm2C − 1

However, it the area is an angle, θ to the field lines then we modify the formula

Φ = EAcosθ

since Acosθ is the component of area to the perpendicular direction


Gauss's Law states that the total electric flux, Φ is given by ;

\phi = \frac{Q}{\epsilon _{0}}

where Q : The total charged enclosed in a gaussian surface

Comparing the definition of electric flux and the statement of Gauss's law we find a convenient way to determine the electric field for any type of surface. It is particularly useful for spherical and cylindrical surfaces where usual methods become troublesome.

\phi =EA= \frac{Q}{\epsilon _{0}}


E= \frac{Q}{\epsilon _{0}A}


Example 1

The electric field due to a spherical conductor


The diagram below shows a charged spherical conductor of radius R. Note that the charges are on the surface of the charged conductor only. ( This is because any charges inside of the conductor would repel each other until they are as far apart from each other as possible)


File:Gauss1.png


i) Electric Field inside the conductor ( r < R )


Consider a spherical Gaussian surface of radius , r <R.

Based on Gauss's Law and the definition of flux;


E= \frac{Q}{\epsilon _{0}A}


Since the charges repel each other , charges are only found on the surface of the charged conductor. Hence Q inside the Gaussian surface is zero.

Hence E inside the charged conductor is also zero .


ii) Electric Field at the surface ( r = R)

Consider a spherical Gaussian surface of radius , r =R

Since the enclosed charge is Q then

E= \frac{Q}{\epsilon _{0}4\pi R^{2}}

by substituting the area of a sphere.


iii) Electric Field outside of the charged conductor ( r>R)

The enclosed charge in a spherical surface of bigger radius than that of the conductor is still Q.

E= \frac{Q}{\epsilon _{0}4\pi r^{2}}


Example 2

The electric field due to a cylindrical conductor

The diagram below shows a cylindrical conductor of radius, R and length L.


File:Gauss2.png

To determine the electric field a distance , r from the cylinder, consider a cylindrical Gaussian surface.

If given that the charge density is λ (C/m) then the total charged enclosed in the Gaussian surface, Q is;

Q = λL

Using the definition of flux and Gauss' Law

E= \frac{Q}{\epsilon _{0}A}

and substituting the surface area of a cylinder


E=\frac{\lambda L}{2\pi rL\epsilon _{0}}


E=\frac{\lambda}{2\pi r\epsilon _{0}}

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