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Electric Fields

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

Source


Source


Representing Fields on Paper

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

\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


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


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.


Millikan’s Oil Drop Experiment

Source



Ngohe/Task

Homework booklet Electric Fields Q2


Akoranga 10 Mahi Tuatahi

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}


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


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:


Whakatika


Task/Ngohe

Working independently, complete Mahi Kāinga/Homework Q6