Question

If an electron moves from one point at a potential of \(-100.0 \mathrm{V}\) to another point at a potential of \(+100.0 \mathrm{V}\) how much work is done by the electric field?

Short answer

Answer: The work done by the electric field on the electron is \(-3.2 \times 10^{-17} \, \mathrm{J}\).

Step-by-step solution

Write down the given information

We are given the initial electric potential, \(V_i = -100.0 \mathrm{V}\), and the final electric potential, \(V_f = +100.0 \mathrm{V}\). We need to find out the work done by the electric field (\(W\)).

Write down the formula to calculate the work done by the electric field

According to the work-energy theorem, the work done by the electric field is equal to the change in the electron's potential energy. The formula for the work done by the electric field is: $$W = q \Delta V$$ where \(q\) is the charge of the electron and \(\Delta V\) is the change in electric potential.

Calculate the change in electric potential

The change in electric potential is the final electric potential minus the initial electric potential: $$\Delta V = V_f - V_i$$ Substitute the given values into the equation: $$\Delta V = (+100.0 \mathrm{V}) - (-100.0 \mathrm{V}) = 200.0 \mathrm{V}$$

Identify the charge of the electron

The charge of the electron, \(q\), is a known constant value: $$q = -1.6 \times 10^{-19} \, \mathrm{C}$$ Note that the charge is negative, as electrons are negatively charged particles.

Calculate the work done by the electric field

Using the formula from Step 2, substitute the values for \(q\) and \(\Delta V\) to find the work done by the electric field: $$W = q \Delta V$$ $$W = (-1.6 \times 10^{-19} \, \mathrm{C}) (200.0 \mathrm{V})$$ Calculate the work: $$W = -3.2 \times 10^{-17} \, \mathrm{J}$$ So, the work done by the electric field on the electron is \(-3.2 \times 10^{-17} \, \mathrm{J}\).