Chapter 16 - Problems - Page 615: 45

We should place the third charge above and to the left of point A, at a distance of $0.254~m$ at an angle of $75.4^{\circ}$ above the horizontal.

First, we can find the components of the electric field at point A due to the two charges at the bottom corners of the square. We can find the distance $r$ from the charge at the opposite corner to point A: $r = \sqrt{(0.300~m)^2+(0.300~m)^2}$ $r = 0.4243~m$ There is a non-zero horizontal component of the electric field due to the charge at the opposite corner from point A. We can find the x-component of the electric field: $E_x = \frac{kq}{r^2}~cos~45^{\circ}$ $E_x = \frac{(9.0\times 10^9~N~m^2/C^2)(7.00\times 10^{-6}~C)}{(0.4243~m)^2}~(cos~45^{\circ})$ $E_x = 2.474\times 10^5~N/C$ The vertical component of the electric field at point A is the sum of the vertical components of the electric field due to each point charge: $E_y = \frac{(9.0\times 10^9~N~m^2/C^2)(7.00\times 10^{-6}~C)}{(0.300~m)^2} + \frac{(9.0\times 10^9~N~m^2/C^2)(7.00\times 10^{-6}~C)}{(0.4243~m)^2}~(sin~45^{\circ})$ $E_y = 9.474\times 10^5~N/C$ We can find the angle $\theta$ above the horizontal that the net electric field points. Note that the net electric field points up and to the left at point A: $tan~\theta = \frac{E_y}{E_x}$ $\theta = tan^{-1}\frac{E_y}{E_x}$ $\theta = tan^{-1}(\frac{ 9.474\times 10^5~N/C}{ 2.474\times 10^5~N/C})$ $\theta = 75.4^{\circ}$ We need to place the third charge so that the electric field due to the third charge is equal in magnitude but opposite in direction to the net electric field due to the two other point charges. The vertical component of the electric field due to the third charge must be equal in magnitude and opposite in direction to $E_y$ above. We can find the distance $d$ from point A that we should place the third charge: $\frac{k~q}{d^2}~sin~\theta = E_y$ $d^2 = \frac{k~q~sin~\theta}{E_y}$ $d = \sqrt{\frac{k~q~sin~\theta}{E_y}}$ $d = \sqrt{\frac{(9.0\times 10^9~N~m^2/C^2)(7.00\times 10^{-6}~C)~(sin~75.4^{\circ})}{9.474\times 10^5~N/C}}$ $d = 0.254~m$ We should place the third charge above and to the left of point A, at a distance of $0.254~m$ at an angle of $75.4^{\circ}$ above the horizontal.