The Electric Field Due to a Point Charge — Question 4

Question 4

One adjustable corner charge cannot generally cancel a two-component field

Charges $+q,+q,-q$ occupy square corners $(-a,-a),(a,-a),(a,a)$, respectively, with $q>0$. An adjustable charge Q occupies $(-a,a)$. Find the field at the center using Coulomb constant k.

1. Derive both center-field components and determine whether any Q makes both zero.

2. Find Q that minimizes the field magnitude and evaluate that minimum.

Question 4 - Solution

One adjustable corner charge cannot generally cancel a two-component field

(a) Method and calculation

$$C=\frac{k}{2\sqrt2a^2},\quad E_x=C(q+Q),\quad E_y=C(3q-Q).$$$$E_x=0\Rightarrow Q=-q,\qquad E_y=0\Rightarrow Q=3q:\quad\text{incompatible}.$$

(b) Interpretation and further calculation

$$E^2=C^2[(q+Q)^2+(3q-Q)^2],\quad\frac{dE^2}{dQ}=4C^2(Q-q).$$$$Q=q,\quad E_{\min}=2\sqrt2Cq=kq/a^2,\quad\frac{d^2E^2}{dQ^2}=4C^2>0.$$The best single adjustment reduces the magnitude but cannot remove both independent components.

Physical check

The common distance from each corner to the center is $\sqrt2a$, not a.