At what distance from a 1.2 x 10\(^{-7}\)C point charge will the electric field intensity be equal to 4.8 x 10\(^{-4}\)NC\(^{-1}\) [ Take \(\frac{1}{4\pi ε_...

Answer Details

The electric field intensity at a distance \(r\) from a point charge obeys an inverse-square law: \[E = \frac{1}{4\pi\varepsilon_0}\cdot\frac{Q}{r^{2}} = \frac{kQ}{r^{2}},\] with \(k = 9.0 \times 10^{9}\ \text{N m}^{2}\text{C}^{-2}\). Since the distance is wanted, make \(r\) the subject: \[r = \sqrt{\frac{kQ}{E}}.\]

Work out the numerator first. \[kQ = (9.0 \times 10^{9})(1.2 \times 10^{-7}) = 1.08 \times 10^{3}\ \text{N m}^{2}\text{C}^{-1}.\] Dividing by the field strength gives \[r^{2} = \frac{1.08 \times 10^{3}}{4.8 \times 10^{-4}} = 2.25 \times 10^{6}\ \text{m}^{2},\] so \[r = \sqrt{2.25 \times 10^{6}} = 1.5 \times 10^{3}\ \text{m} = 1.5\ \text{km}.\] The field intensity falls to \(4.8 \times 10^{-4}\ \text{N C}^{-1}\) at 1.5 km from the charge.

The commonest error is forgetting the square root and quoting \(2.25 \times 10^{6}\), or taking the root of only part of the expression. Handle the powers of ten deliberately: to take the square root of a number in standard form, first arrange the index to be even, as with \(2.25 \times 10^{6}\), so that halving it gives \(10^{3}\) exactly. Because the relationship is inverse-square, notice also that reducing the field to a quarter of a value doubles the distance, and the final answer had to be converted from metres to kilometres to match the way the alternatives are written.