Question

How much work would be done by an electric field in moving a proton from a point at a potential of \(+180 . \mathrm{V}\) to a point at a potential of \(-60.0 \mathrm{~V} ?\)

Short answer

Question: Calculate the work done by an electric field in moving a proton from a point at a potential of +180 V to a point at a potential of -60 V. Answer: The work done by the electric field is \(3.84\times 10^{-17}\) J.

Step-by-step solution

Identify the given values

In this problem, we have the following given values: - Potential at the initial point (V1): \(+180\) V - Potential at the final point (V2): \(-60\) V - Charge of a proton (q): \(+1.6\times10^{-19}\) C

Calculate the potential difference

To find the potential difference, subtract the final potential (V2) from the initial potential (V1): \(\Delta V = V_1 - V_2 = (+180) - (-60) = 240\) V

Calculate the work done

Now use the formula \(W = q\Delta V\) to find the work done by the electric field in moving the proton: \(W = (1.6\times10^{-19}\,\text{C})(240\,\text{V}) = 3.84 \times 10^{-17}\,\text{J}\)

Write the final answer

The work done by the electric field in moving a proton from a point at a potential of \(+180\) V to a point at a potential of \(-60\) V is \(3.84\times 10^{-17}\) J.