**Physics, SS 3 Week 3**

**Topic:** **ELECTRIC FIELD **

**Coulombs Law**

The electric force between two point charges is directly proportional to the magnitude of each charge (q_{1}, q_{2}), inversely proportional to square of the separation between their centers (r), directed along the separation vector connecting their centers.

This relationship is known as Coulomb’s Law. Charles Augustin de Coulomb (1736-1806) France. As an equation it is usually written in one of two forms …

F = k_{e} q_{1}q_{2}/r^{2} or F = 1/4πe_{o}. q_{1}q_{2}/r^{2}

Electrostatic constant k_{e }= 8.99 x 10^{9 }Nm/C^{2},

Vacuum Permittivity e_{o} = 8.85 x 10^{-12}C^{2}/Nm^{2}

When two charges have the same sign their product is positive, which means the force vector is directed with the separation vector and the action is repulsive.

When two charges have the opposite sign their product is negative, which means the force vector is directed against the separation vector and the action is attractive.

The electrical force, like all forces, is typically expressed using the unit Newton. Being a force, the strength of the electrical interaction is a vector quantity that has both magnitude and direction. The direction of the electrical force is dependent upon whether the charged objects are charged with like charge or opposite charge and upon their spatial orientation. By knowing the type of charge on the two objects, the direction of the force on either one of them can be determined with a little reasoning. In the diagram below, objects A and B have like charge causing them to repel each other. Thus, the force on object A is directed leftward (away from B) and the force on object B is directed rightward (away from A). On the other hand, objects C and D have opposite charge causing them to attract each other. Thus, the force on object C is directed rightward (toward object D) and the force on object D is directed leftward (toward object C). When it comes to the electrical force vector, perhaps the best way to determine the direction of it is to apply the fundamental rules of charge interaction (opposites attract and likes repel) using a little reasoning.

**Example**

Suppose that two point charges, each with a charge of +1.00 Coulomb are separated by a distance of 1.00 meter. Determine the magnitude of the electrical force of repulsion between them.

Q_{1} = 1.00 C, Q_{2 }= 1.00 C, d = 1.00 m, F_{elect} =?

F = 1/4πe_{o}. q_{1}q_{2}/r^{2}

= (9.0 x 10^{9} Nm^{2}/C^{2}) x (1.00 C) x (1.00 C) / (1.00 m)^{2}

= 9.0 x 109 N

**Electric Field Intensity**

Electric field strength is a vector quantity; it has both magnitude and direction. The magnitude of the electric field strength is defined in terms of how it is measured. Let’s suppose that an electric charge can be denoted by the symbol Q. This electric charge creates an electric field; since Q is the source of the electric field, we will refer to it as the **source charge**. The strength of the source charge’s electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a testcharge since it is used to *test* the field strength. The test charge has a quantity of charge denoted by the symbol q. When placed within the electric field, the test charge will experience an electric force – either attractive or repulsive. As is usually the case, this force will be denoted by the symbol F. The magnitude of the electric field is simply defined as the force per charge on the test charge.

Note that in the figure below d indicates the distance between the source charge and the test charge, which could also be denoted by ‘r’.

Electric Field Strength = Force/Charge

If the electric field strength is denoted by the symbol E, then the equation can be rewritten in symbolic form as

E = F/q, F = qE, Electric Force = Electric Charge x Electric Field

We can obtain the equation for the Field Intensity E due to a point charge q at a distance r from that charge making use of these two equations E = F/q and F = qE we assume the point charge is situated vacuum

F = 1/4πe_{o}. Qq/r^{2}

E = F/q = 1/4πe_{o} Qq/r^{2}q = Q/4πe_{o} r^{2}

Hence we can write

E= Q/4πe_{o} r^{2}

This is the field intensity due to a charge Q at a distance r from the charge.

**Electric Potential**

In a gravitational field, forces of attraction acts between masses. Therefore in order to separate two masses, work must be done against the field.

Similarly, in electrostatic field where forces of attraction or repulsion act between charges, work must be done against this field in order to move charges against the field.

The gravitational potential at any point in a gravitational field is the work done per unit in bringing mass to that point from an arbitrarily chosen zero level of the earth. Similarly in an electrostatic or electric field the electric potential at a point is described as the work done per unit positive charge in bringing charge to that point from an arbitrarily chosen zero of potential. If work is done against the field, the potential is positive. If work is done by the field, the potential is negative. Points of positive potentials are said to be at a higher potential than those of negative potential. Thus in an electric cell, the positive terminal is at a higher potential than the negative terminal. The arbitrarily chosen zero of potential is taken at infinity, but in practical terms, ** the earth is taken to be at zero potential**. Thus whenever a conducting body is connected to the earth it is said to be

**or to have a zero potential.**

*‘grounded’ or ‘earthed’*Electric Potential (V) at a point is defined as the work done in bringing unit positive charge from infinity to that point against the electrical forces of the field.

Potential difference (V_{AB}) between two points A and B is the work done in taking unit positive charge from one point to the other in the electric field.

Both the potential and the potential difference are scalar quantities having the dimension of work/charge. The unit is the volt.

1 volt = 1 Joule/Coulomb

The potential difference between two points is one volt if the work done in taking one Coulomb of positive charge from one point to the other is one joule.

*W* = *qV*

(Joules) = (Coulomb x Volts) or *V = W/q*

Electric intensity is related to the electric potential difference between two points through the equation.

** E = v/d**, where d is the distance between two points in an electric field

Hence we have, ** V =Ed**,volt = NC

^{-1}m

The unit of ** E = v/d** can also be in volt per meter but we have E = q/4πe

_{o}d

^{2 }or q/4πe

_{0}d

^{2}

Putting this in ** V =Ed**, we have

*v = q/4π**e _{o}d^{2}* x

*d*or

*v = q/4π*

*e*

_{o}dWhere v is the potential at a point due to a charge q at a distance d from the charge.

Two parallel plates are charged to a voltage of 40V. If they are separated by a distance of 10.0 cm, calculate the electric intensity between them.

Solution

E = V/d

= 40 (volt)/0.1 m

= 400 Vm^{-1}

**Capacitors and Capacitance**

Capacitors

A capacitor is an electronic device for storing charge. Capacitors can be found in almost all but the most simple electronic circuits. There are many different types of capacitor but they all work in essentially the same way. A simplified view of a capacitor is a pair of metal plates separated by a gap in which there is an insulating material known as the dielectric. This simplified capacitor is also chosen as the electronic circuit symbol for a capacitor is a pair of parallel plates as shown below

The symbol above is for an unpolarised capacitor

*On the picture are different capacitors look*.

**Arrangement of Capacitors – Series and Parallel**

**Series**

Consider the series network of capacitors shown in the figure above where the positive plate is connected to the negative plate of the next. What is the equivalent capacitance of the network? Look at the plates in the middle, these plates are physically disconnected from the circuit so the total charge on them must remain constant. It follows that when a voltage is applied across both of the capacitors, the charge +*Q* on the positive plate of capacitor *C*_{1} must be balanced by the charge –*Q* on the negative plate of capacitor *C*_{2}. The net result is that both capacitors possess the same charge Q. The potential drops *V*_{1} and *V*_{2} across the two capacitors are in general, different. However, the sum of these drops equals the total potential drop *V* applied across the input and output wires. *V*=*V*_{1} + *V*_{2}. The equivalent capacitance of the pair is again *C*_{T}=*Q*/*V*. Thus, 1/*C*_{T} = *V*/*Q* = (*V*_{1} + *V*_{2})/*Q* = *V*_{1}/*Q* + *V*_{2}/*Q* giving

By connecting capacitors in series you store less charge so does ever make sense to connect capacitors in series? It is sometimes done because capacitors have maximum working voltages, and with two 900 volt maximum capacitors in series, you can increase the working voltage to 1800 volts.

**Parallel**

For a parallel circuit such as in Figure above, the voltages are the same across each component. However the total charge is divided between the two capacitors since it must distribute itself such that the voltage across the two is the same. Also, since the capacitors may have different capacitances *C*_{1} and *C*_{2} the charges *Q*_{1} and *Q*_{2} must also be different. The equivalent capacitance *C*_{T} of the pair of capacitors is simply the ratio *Q/V* where *Q*=*Q*_{1}+*Q*_{2} is the total stored charge. It follows that *C*_{T} = *Q*/*V* = (Q_{1}+Q_{2})/*V* = *Q*_{1}/*V* + *Q*_{2}/*V* giving

1/C_{t }= 1/C_{1} + 1/C_{2}

C_{T = }C_{1} + C_{2}

The overall capacitance increases by adding together capacitors in parallel so we create larger capacitances than is possible using a single capacitor.

### Capacitor – Energy Stored

The energy stored in a capacitor can be expressed as

W = 1/2 CV^{2}

Where W = energy stored (Joules), C = capacitance (Farad), V = potential difference (Voltage)

**Questions**

1. Two balloons are charged with an identical quantity and type of charge: -6.25 nC. They are held apart at a separation distance of 61.7 cm. Determine the magnitude of the electrical force of repulsion between them.

A. 8.0 x 10^{-6} N B. 9.23 x 10^{-7} N C. 7.54 x 10^{-8 }N D. 54 x 10^{-5 }N

2. Two balloons with charges of +3.37 µC and -8.21 µC attract each other with a force of 0.0626 Newton. Determine the separation distance between the two balloons.

A. +1.99 m B -1.99 m C. 7.34 m D. 2.65 m

3. The energy stored in a *10 μF* capacitor charged to *230 V* can be calculated as

A. 1.46 J B. 0.26 J C. 8 J D. 3.45 J

4. Electric field strength has

A. Only direction B Only Magnitude C. Magnitude and direction D. None of the options

5. By connecting capacitors in series you store

A. More charges B. No Charge C. Average charges D. Less charge

Questions

1. B 2. A 3. B 4. C 5. D

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