Question

Suppose a uniform electric field of magnitude \(100.0 \mathrm{N} / \mathrm{C}\) exists in a region of space. How far apart are a pair of equipotential surfaces whose potentials differ by \(1.0 \mathrm{V} ?\)

Short answer

Answer: The distance between the equipotential surfaces is \(0.01 \mathrm{m}\).

Step-by-step solution

Write down the given values

The electric field \(E = 100.0 \mathrm{N} / \mathrm{C}\) and the potential difference \(\Delta V = 1.0 \mathrm{V}\) are given.

Write down the formula for electric field and potential difference

The formula to find the distance between equipotential surfaces is \(E = \frac{\Delta V}{d}\).

Rearrange the formula to find the distance

We can rearrange the formula to find the distance between equipotential surfaces as follows: \(d = \frac{\Delta V}{E}\).

Substitute the given values into the formula

Now, we can substitute the given values of electric field and potential difference into the formula to find the distance: \(d = \frac{1.0 \mathrm{V}}{100.0 \mathrm{N} / \mathrm{C}}\).

Calculate the distance

By dividing the potential difference by the electric field, we find the distance: \(d = \frac{1.0 \mathrm{V}}{100.0 \mathrm{N} / \mathrm{C}} = 0.01 \mathrm{m}\). The equipotential surfaces whose potentials differ by \(1.0 \mathrm{V}\) are \(0.01 \mathrm{m}\) apart.