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Potential Energy and Voltage

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
  4. Worksheet 1: Q6a, 6bb, Q11a, 11b, 11d
  5. Textbook: Electric Fields Q7-8 pg. 192

Whakatika 🔗

  1. An area of influence in where a charged object will feel a force

$$ \begin{aligned} E &= \frac{F}{q} \newline 3000 &= \frac{F}{25 \times 10^{-6}} \newline F &= 3000 \times 25 \times 10^{-6} \newline F &= 0.075N \end{aligned} $$

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 similar 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

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

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.

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

Pātai 🔗

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)

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

(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 $F=Eq$ 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)

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

(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.