Question

A point charge of \[1\ \mu\text{C}\] produces a potential \[V = \frac{kq}{r}\]. At what distance from the charge does the potential equal 90 V? (Take \[k = 9 \times 10^9\ \text{N m}^2/\text{C}^2\])

Options

• 50 m

• 100 m

• 900 m

• 10 m

Answer 1

A point charge's potential falls off as \[V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} = \frac{kq}{r}\]. Rearranged: \[r = \frac{kq}{V} = \frac{9 \times 10^9 \times 1 \times 10^{-6}}{90} = 100\ \text{m}\]. The 90 V equipotential sphere around this 1 μC charge has a radius of 100 m — every point on that sphere is at exactly 90 V.