Electric Fields

12PHYS - Electricity

Finn LeSueur

2021

Akoranga 4 Mahi Tuatahi

  1. If a balloon has charge of \(-0.5C\): did the balloon lose or gain electrons, and how many?
  2. If a Van der Graaff generator has charge of \(4C\) did it lose or gain electrons, and how many?

Whakatika

  1. If a balloon has charge of \(-0.5C\): did the balloon lose or gain electrons, and how many?
\[\begin{aligned} & \text{Negative C means electrons are gained (negative charge)} \newline num_{e} &= \frac{-0.5}{-1.6 \times 10^{-19}} \newline num_{e} &= 3.125 \times 10^{18} && \text{ electrons gained} \end{aligned}\]
  1. If a Van der Graaff generator has charge of \(4C\) did it lose or gain electrons, and how many?
\[\begin{aligned} & \text{Positive C means electrons are lost (positive charge)} \newline num_{e} &= \frac{4}{1.6 \times 10^{-19}} \newline num_{e} &= 2.5 \times 10^{19} && \text{ electrons lost} \end{aligned}\]

Ngā Whāinga Ako

  1. Be able to draw electric fields
  2. Be able to determine which way a charge will move in a field

Write the date and ngā whāinga ako in your book

Electric Fields

An electric field is an area of influence in which a charged object will feel a force

  • A field can exist in a vacuum or inside an substance.
  • An electric field exists in the space around a charged objects e.g. Van der Graaf generator, electrons, protons or a charged balloon.
Source
  • The closer you are to a field source (e.g. the charged particle), the greater the force exerted by the field
  • The field extends out to infinity, but the strength diminishes exponentially
Source

Representing Fields on Paper

  • We draw electric fields as a series of arrows, which indicate direction
  • By convention the fields goes from positive to negative
  • Lines close together indicate a strong field
  • Lines far apart indicate a weak field
  • We can see the lines are closer together near the charge, indicating a stronger field!
Source

Tūhura/Investigation

Add six boxes to your notes labelled:

  1. Positive Charge
  2. Negative Charge
  3. Two Positive Charges
  4. Two Opposite Charges
  5. Two positively charged parallel plates
  6. Two oppositely charged parallel plates

Use the simulation on Google Classroom to help you discover the fields for the first four, and use your intuition to discover 5-6.

Akoranga 5 Ngā Whāinga Ako

  1. Be able to relate \(\vec{E}, F\) and \(q\).
  2. Be able to calculate electric field strength
  3. Be able to calculate electric field potential difference

Write the date and nga whāinga ako in your book

Electric Fields & Forces

  • Electric fields exert a force on charged objects & particles.
  • Charged particles are simultaneously repelled by their similarly charged terminal, and attracted by their oppositely charged terminal.
\[\begin{aligned} \vec{E} &= \frac{\vec{F}}{q} \newline \text{Electric Field Strength} &= \frac{Force}{Charge} \newline \end{aligned}\]

Therefore \(\vec{E}\) has units of \(\frac{N}{C}\) or \(NC^{-1}\)

Pātai

Is electric field strength a scalar or a vector?

Whakatika

It is a vector, because one of the components that creates it is force, and force has both direction and magnitude. Because it is a vector we can give it the notation \(\vec{E}\). You can use this notation with any other vector, too.

\(\vec{E}\) Pātai Tahi

A small charge of \(2 \times 10^{-4}C\) experiences a force of \(1.5\times10^{-4}N\). Calculate the electric field strength.

\[\begin{aligned} & && \text{(K)} \newline & && \text{(U)} \newline & && \text{(F)} \newline & && \text{(S+S)} \end{aligned}\]

\(\vec{E}\) Pātai Tahi: Whakatika

A small charge of \(2\times10^{-4}C\) experiences a force of \(1.5\times10^{-4}N\). Calculate the electric field strength.

\[\begin{aligned} & \vec{E} = \frac{\vec{F}}{q} \newline & \vec{E} = \frac{1.5\times10^{-4}}{2\times10^{-4}} \newline & \vec{E} = 0.75NC^{-1} \end{aligned}\]

Task/Ngohe

  • Textbook: Electric Fields Q1, Q2, Q4, Q5
    • New: pg. 192
    • Old: pg. 177
  • Homework Booklet: Q3a-b

Akoranga 6 Mahi Tuatahi

A field of strength \(2NC^{-1}\) is acting with a force of \(0.5N\) on a charged particle. The field exists between two parallel plates.

  1. Draw a diagram illustrating the system.
  2. Calculate the charge of the particle. Indicate the force arrrow on your diagram.
\[\begin{aligned} & && \text{(K)} \newline & && \text{(U)} \newline & && \text{(F)} \newline & && \text{(S+S)} \end{aligned}\]

Te Whāinga Ako

  1. Be able to calculate the voltage between parallel plates

Write the date and te whāinga ako in your book

Voltage between Plates

\[\begin{aligned} & \vec{E} = \frac{V}{d} \newline & \vec{E} = \text{ electric field strength} \newline & V = \text{ voltage applied on the two plates} \newline & d = \text{ distance between the two plates} \end{aligned}\]

Pātai

An object with charge \(-4 \mu C\) is placed between two charged plates \(2cm\) apart with a potential difference of \(500V\).

  1. Calculate the electric field strength between the two charged plates
  2. Calculate the size of the force experienced by the charged object
Source

Whakatika

  1. Calculate the electric field strength between the two charged plates
\[\begin{aligned} & \vec{E} = \frac{\vec{V}}{d} \newline & \vec{E} = \frac{500}{0.02} \newline & \vec{E} = 25000NC^{-1} \end{aligned}\]
  1. Calculate the size of the force experienced by the charged object
\[\begin{aligned} & \vec{F} = \vec{E}q \newline & \vec{F} = 25000 \times -4 \times 10^{-6} \newline & \vec{F} = -0.1N \end{aligned}\]

Tūhura/Investigation

Setup:

  1. Open the link on Google Classroom
  2. Choose “Capacitance”
  3. Uncheck “Capacitance”, “Top Plate Charge”, “Bar Graphs” and “Stored Energy”
  4. Check “Plate Charges”, “Electric Field”, and “Current Direction”
  5. Attach a voltmeter to the wire on either side of the parallel plates

In Your Books:

  1. What do you observe about the amount of charge on each plate as you vary the battery from 1.5V to -1.5V?
  2. Similarly, what happens to the field strength (as indicated by the number of lines)
  3. And, what happens when you vary the plate separation at a constant voltage?

Task/Ngohe

  • Worksheet 1: Q6a-b, Q11a-b,d
  • Textbook: Electric Fields Q7-8
    • New: pg. 192
    • Old: pg. 177

Akoranga 8 Mahi Tuatahi

An object with charge \(25 \times 10^{-6}C\) is placed in an electric field with strength \(3000NC^{-1}\).

  1. Define electric field
  2. Calculate the force experienced by the object inside the electric field, and indicate direction
  3. Draw a diagram illustrating the situation

Whakatika

  1. An area of influence in where a charged object will feel a force
  2. \[\begin{aligned} & \vec{E} = \frac{\vec{F}}{q} \newline & 3000 = \frac{\vec{F}}{25 \times 10^{-6}} \newline & \vec{F} = 3000 \times 25 \times 10^{-6} \newline & \vec{F} = 0.075N \end{aligned}\]

Ngā Whāinga Ako

  1. Be able to relate electric potential energy and voltage
  2. Use work in an electric field context

Write the date and ngā whāinga ako in your book

Electric Potential Energy

Electric potential energy is akin to gravitational potential energy. Moving a charged particle against the direction of the field is similar to lifting an object up in a gravitational potential field.

Therefore, different points in an electric field will have different potential energies

Gravity Field:

\[\begin{aligned} & W = F \times d \newline & W = mg \times h \newline & W = mgh = E_{p-grav} \end{aligned}\]

Electric Field:

\[\begin{aligned} & W = F \times d \newline & W = Eq \times d \newline & W = Eqd = E_{p-elec} \end{aligned}\]

Voltage

The difference in electric potential energy between two points

OR

The work needed per unit charge to move a test charge between two points.

\[\begin{aligned} V &= \frac{E_{p}}{q} \newline \text{Field Voltage} &= \frac{\text{Potential Energy}}{\text{Charge}} \end{aligned}\]

Mahi Tuatahi Continued

An object with charge \(-25 \times 10^{-6}C\) is placed in an electric field with strength \(3000NC^{-1}\).

  1. In the object moves against the field by \(2m\), does it gain or lose electric potential energy? And how much?

Whakatika

Because it moves against the field, it gains electric potential energy.

\[\begin{aligned} & E_{p} = Eqd \newline & E_{p} = 3000 \times 25 \times 10^{-6} \times 2 \newline & E_{p} = 0.15J \end{aligned}\]

Akoranga 9 Mahi Tuatahi

Homework booklet Electric Fields Question 4

Whakatika

Hamish connects a circuit comprised of a 6.0 V battery, 1.0 m of Nichrome resistance wire and two connecting wires. The battery produces a uniform electric field in the Nichrome resistance wire. Assume that the connecting wires have no resistance.

(a) Calculate the strength of the electric field in the Nichrome resistance wire. (A)

\[\begin{aligned} & V=6V, d=1m && \text{(K)} \newline & \vec{E}=? && \text{(U)} \newline & \vec{E} = \frac{V}{d} && \text{(F)} \newline & \vec{E} = \frac{6}{1} = 6Vm^{-1} && \text{(S+S)} \end{aligned}\]

(b) Explain what happens to the size of the electric force on an electron as it travels through the Nichrome resistance wire. (M)

Throughout the wire the field strength is constant, and the charge on the electron is also constant. Therefore using \(F=\vec{E}q\) we can see that the force is also constant.

(c) Calculate the distance moved by an electron as it loses 9.6 × 10-20 J of electrical potential energy. (M)

\[\begin{aligned} & W = 9.6\times10^{-20}J, \vec{E} = 6Vm^{-1}, q=-1.6\times10^{-19}&& \text{(K)} \newline & d = ? && \text{(U)} \newline & W = Fd = Eqd && \text{(F)} \newline & 9.6\times10^{-20} = 6 \times (-1.6\times10^{-19}) \times d && \text{(S+S)} \newline & d = \frac{9.6\times10^{-20}}{6 \times (-1.6\times10^{-19})} = 0.1m \end{aligned}\]

(d) Hamish then adds another 6.0 V battery in series AND shortens the wire to 0.50 m. Write a comprehensive explanation on what will happen to the size of the force on the electron. (E) Calculations are not needed.

  • Adding another \(6V\) battery will double the electric field strength according to \(\vec{E} = {V}{d}\).
  • Shortening the wire from \(1m\) to \(0.5m\) will also double the field strength according to \(\vec{E} = {V}{d}\).
  • Both of these factors will result in a 4x stronger electric field.
  • Assuming the charge of the electron stays constant, according to \(F=\vec{E}q\), the force experienced will also be 4x stronger.

Millikan’s Oil Drop Experiment

Source
  • In the early 1900s we did not know what the charge of the electron was
  • Millikan figured that if the force exerted by an electric field was strong enough, it could balance the weight force due to gravity
  • By measuring the field strength needed to balance many different droplets of charged oil, he figured out that the charges were multiples of \(1.5924(17)\times10^{−19}C\)!

Ngohe/Task

Homework booklet Electric Fields Q2

Akoranga 10 Mahi Tuatahi

  • In an X-ray machine, a heating element releases electrons from a negatively charged plate called the cathode. The electrons are then accelerated by an electric field that exists between the cathode and a positively charged tungsten plate called the anode.
  • The cathode and the anode are connected to a high voltage source of \(20 000 V\). The distance between the cathode and anode plates is \(0.050 m\). The beam of electrons causes X-rays to be released from the anode.
  • Charge on an electron = \(-1.60\times10^{–19}C\)
  • Mass of an electron = \(9.11\times10^{–31}kg\)

a) Calculate the electric field strength between the plates, and state its direction. (A)

Whakatika

\[\begin{aligned} & V=20000V, d=0.05m && \text{(K)} \newline & \vec{E} = ? && \text{(U)} \newline & \vec{E} = \frac{V}{d} && \text{(F)} \newline & \vec{E} = \frac{20000V}{0.05m} = 400000Vm^{-1} && \text{(S+S)} \newline & \text{Direction: from anode (+ve) to cathode (-ve)} \end{aligned}\]
  • In an X-ray machine, a heating element releases electrons from a negatively charged plate called the cathode. The electrons are then accelerated by an electric field that exists between the cathode and a positively charged tungsten plate called the anode.
  • The cathode and the anode are connected to a high voltage source of \(20 000 V\). The distance between the cathode and anode plates is \(0.050 m\). The beam of electrons causes X-rays to be released from the anode.
  • Charge on an electron = \(-1.60\times10^{–19}C\)
  • Mass of an electron = \(9.11\times10^{–31}kg\)

b) State what type of energy an electron would have at the cathode (negative plate), and what would happen to that energy as the electron moved towards the anode (positive plate). (M)

Whakatika

  • Particle Point of View:
    • Maximum electric potential energy when on the cathode.
    • This electric potential energy is transformed to kinetic energy as it moves away from the cathode.
  • Field Point of View:
    • The field is doing work on the particle, so it loses electric potential energy.
    • This energy is transferred to the particle in the form of kinetic energy
  • In an X-ray machine, a heating element releases electrons from a negatively charged plate called the cathode. The electrons are then accelerated by an electric field that exists between the cathode and a positively charged tungsten plate called the anode.
  • The cathode and the anode are connected to a high voltage source of \(20 000 V\). The distance between the cathode and anode plates is \(0.050 m\). The beam of electrons causes X-rays to be released from the anode.
  • Charge on an electron = \(-1.60\times10^{–19}C\)
  • Mass of an electron = \(9.11\times10^{–31}kg\)

c) Calculate the speed of the electron as it reaches the anode (positive plate). (M)

Whakatika

The field does work on the particle, so it loses energy:

\[\begin{aligned} & \vec{E}=400000Vm^{-1}, q=-1.6\times10^{-19}C, d=0.05m && \text{(K)} \newline & E_{p}=W=? && \text{(U)} \newline & E_{p} = W = Fd = \vec{E}qd && \text{(F)} \newline & W = 400000Vm^{-1} \times (-1.6\times10^{-19}C) \times 0.05m = -3.2\times10^{-15}J && \text{(S+S)} \end{aligned}\]

Assuming no friction, all \(E_{p}\) converted to \(E_{k}\):

\[\begin{aligned} E_{p} &= E_{k} \newline -3.2\times10^{-15}J &= \frac{1}{2}mv^{2} \newline v &= \sqrt{\frac{2\times(-3.2\times10^{-15})}{9.11\times10^{-31}}} = 8.39\times10^{7}ms^{-1} \end{aligned}\]

In 1909 Millikan used two oppositely charged metal plates to keep a charged oil drop falling at terminal velocity when he was experimenting to find the charge of an electron. A modified form of his experiment keeps an oil drop stationary.

d) Discuss how it was possible to make the oil drop stationary between the plates. (E)

In your comprehensive answer you should:

  • identify the forces acting on the oil drop
  • describe how the forces can combine to cause the oil drop to be stationary
  • explain what type of charge the oil drop must have in order to remain stationary.

Whakatika

  • The forces acting on the oil drop are downward weight force and upward electrical force.
  • These two forces must be balanced, as the oil drop is stationary.
  • For the electrical force to be upwards, the type of charge on the oil drop must be opposite to the charge on the top plate.

Task/Ngohe

Working independently, complete Mahi Kāinga/Homework Q6