Example

Question:

Four charges are arranged at the corners of a square \(ABCD\) of side \(d\), as shown in Fig. 2.15(a).
(a) Find the work required to put together this arrangement.
(b) A charge \(q_0\) is brought to the centre \(E\) of the square, the four charges being held fixed at its corners. How much extra work is needed to do this?

Solution:

(a) Since the work done depends on the final arrangement, we calculate work needed for one way of adding the charges at \(A, B, C, D\):
(i) Work needed to bring \(+q\) to \(A\): 0.
(ii) Work to bring \(-q\) to \(B\) when \(+q\) at \(A\): \[ -q \times \left(\frac{q}{4\pi\varepsilon_0 d}\right) = -\frac{q^2}{4\pi\varepsilon_0 d} \] (iii) Work to bring \(+q\) to \(C\) when \(+q\) at \(A\), \(-q\) at \(B\): \[ +q \left( \frac{q}{4\pi\varepsilon_0 d\sqrt{2}} - \frac{q}{4\pi\varepsilon_0 d} \right) = -\frac{q^2}{4\pi\varepsilon_0 d}(1-\frac{1}{\sqrt{2}}) \] (iv) Work to bring \(-q\) to \(D\) after other three charges: \[ -q\left(\frac{q}{4\pi\varepsilon_0 d} - \frac{q}{4\pi\varepsilon_0 d\sqrt{2}} + \frac{q}{4\pi\varepsilon_0 d} \right) = -\frac{q^2}{4\pi\varepsilon_0 d}\left(2-\frac{1}{\sqrt{2}}\right) \] Add up all steps: \[ = -\frac{q^2}{4\pi\varepsilon_0 d}\left[0 + 1-\frac{1}{\sqrt{2}} + 2-\frac{1}{\sqrt{2}}\right] = -\frac{q^2}{4\pi\varepsilon_0 d}\left(4-\sqrt{2}\right) \] (b) The extra work to bring a charge \(q_0\) to centre \(E\) when all corner charges are fixed is \(q_0 \times\) (potential at \(E\)), but potential there is zero due to symmetry, so the extra work required is zero.