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.

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

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

  Applicable in vacuum or in air.

This applies in vacuum or free space since $ε 0$ 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, $ε$

   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

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

$F R e s u l t a n t = F a - F b = 4.5 k N$ ( to the right)

Example 3 Resultant force due to charges at angles

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

Qa = +3 $μ$ C

Qb = +4 $μ$ C

Qc = -5 $μ$ C

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

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.

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

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

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.

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

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

$X = 67.4 0$

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.

$E=\frac{F}{q}$ ( $N C - 1$ or $V m - 1$ )

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.

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.

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.

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 = $45 o$ 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 : $N m 2 C - 1$

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)

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.

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

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

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

Electric Potential

Electric potential , V (v)

Electric potential at a point due to a point charge, Q

$V=\frac{Q}{4\pi \epsilon r}$

The electric potential, V at a point is defined as the work done to bring a unit positive charge from infinity to that point.

Unit : Volts (V)

The following discussion is a simplified explanation of how the formula comes about.

Remember that the electric field lines point radially out of a point source. The lines indicate the path a free positive charge would take if released into the electric field. If a charge is brought against the electric field then the electrostatic force, F would oppose the motion.

We know that work,W is the product of force,F and displacement, r.

For small displacements in the opposite direction to the electric field, -dx work is done, dW.

dW= F (-dx)

The force involved is the elctrostatics force

$F=\frac{q_{1}q_{2}}{4\pi \epsilon r^{2}}$

The total work done to bring a charge from infinity to the point we are referrring to is found by integrating work with respect to the displacement giving

$W=\frac{qQ}{4\pi \epsilon r}$

Since potential difference is defined as the work per unit charge then electric potential at a point due to a point charge Q is

$V=\frac{Q}{4\pi \epsilon r}$

Q : The source charge

r: the displacement between the source charge and the point of reference

Electric potential of a sphere

If we charge a sphere , the charges will accumulate only on the surface of the sphere. This is because any charge inside of a conductor will repel each other as discussed in "Electric Field".

The variation of electric potential with distance from the centre of the sphere is shown below.

Electric potential energy, U

The electric potential energy, U is related to the potential

$U = q V$

Hence substituting for potential gives

$U = \frac{qQ}{4\pi \epsilon r}$

Electric Potential difference, V

Consider two points at difference distances away from a source charge. The potential at these points are difference. If a charge is to be brought towards the source, work has to be done .

The work done to bring a unit positive charge from one point to the other against the electric field is given by the difference in potential between those two points. With reference to the figure above, the potential difference between point A and B is Va - Vb.

Example:

The potential difference between point A and point B is 200 V.

Determine the work done in bringing a 0.1C charge from point B to point A against t electric field.

Potential difference is the work done in bring a unit ( which means a +1C ) positive charge between the two points against the electric field. For 0.1C the work done,

W = qV

W = 0.1 (200)

W= 20J

The relationship between electric field strength, E and electric potential, V

'	Electric Field Strength, E	Electric Potential,V
Formula	$E=\frac{Q}{4\pi \epsilon r^{2}}$	$V=\frac{Q}{4\pi \epsilon r}$
Relation	$E=-\frac{\mathrm{d} V}{\mathrm{d} x}$	$V=-\int_{\infty }^{r}E\, dx$
Graph

Equipotential surfaces

As the name suggest equipotential surfaces are surfaces of equal potential. Every point on an equipotential surface has the same value of electric potential. Since all the points are at equal potential then no work is done in moving a charge between points on the same equipotential surface.

Example

The potential around a charged sphere decreases as the displacement increases. Equipotential surfaces are at right angles to the electric field hence are radial from the center of the sphere.

Point X and Y are on the same equipotential surface so the potential at X and Y is the same which is 1000V. Moving a positive unit charge from X to Y will not require any work.

Point Y and point C are at difference potentials. Moving a positive unit charge from Y to C will require work amounting to the potential difference

i.e Work done = 1 C(1600-1000) J

If the charge is not a unit charge say q = 0.1C then the work done becomes

W=qV

q : Charge

V : potential difference

W=0.1(600)

W= 60J

Electrostatics

From StpmWiki

Contents

Coulomb's Law

Electric Field

Gauss's Law

Electric Potential

Equipotential surfaces

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