Question

section 24.1 electric potential and potential difference

1. how much work is done (by a battery, generator, or some other source of potential difference) in moving avogadros number of electrons from an initial point where the electric potential is \\(9.00\text{ v}\\) to a point where the electric potential is \\(-5.00\text{ v}\\)? (the potential in each case is measured relative to a common reference point.)

Explanation:

Identify given physical quantities

We are given:

• Number of charges: Avogadro's number of electrons, \(N = 6.022 \times 10^{23}\).
• Charge of one electron: \(q_e = -1.602 \times 10^{-19}\text{ C}\).
• Initial potential: \(V_i = 9.00\text{ V}\).
• Final potential: \(V_f = -5.00\text{ V}\).

Calculate the total charge moved

Using the Electrostatics concept, we find the total charge \(q\):

$$ q = N \cdot q_e = (6.022 \times 10^{23}) \cdot (-1.602 \times 10^{-19}\text{ C}) \approx -9.647 \times 10^4\text{ C} $$

Determine the potential difference

The change in electric potential \(\Delta V\) is:

$$ \Delta V = V_f - V_i = -5.00\text{ V} - 9.00\text{ V} = -14.00\text{ V} $$

Calculate the work done by the source

The work done by the external source (battery/generator) to move the charge is:

$$ W = q \Delta V $$

$$ W = (-9.647 \times 10^4\text{ C}) \cdot (-14.00\text{ V}) \approx 1.35 \times 10^6\text{ J} $$

Answer:

\(1.35 \times 10^6\text{ J}\) (or \(1.35\text{ MJ}\))