Question

At the corners A, B and C of a square ABCD (each side 0.2 m) positive charges of 2$\times {10}^{-9}$ c, 4$\times {10}^{-9}$ C and 6$\times {10}^{-9}$ C are placed. What is the intensity of the electric field at D?
(1836 N/C at 24 43' with CD produced)

Answer 1

The charge is placed as shown in the figure.

$2\times {10}^{-9}atpositionA,4\times {10}^{-9}atpositionBand6\times {10}^{-9}$ at C

$E_A,E_{B}and{E}_C$is produce due to charge present at position A ,B and C

$\Rightarrow \overrightarrow{E_A}=\frac{-K{q}_A}{d^2}\hat{j}=-\frac{9\times {10}^9\times 2\times {10}^{-9}}{(0.2{)}^2}\hat{j}=-50\hat{j}N/C.$

$\Rightarrow \overrightarrow{E_C}=\frac{-K{q}_c}{d^2}\hat{i}=\frac{9\times {10}^9\times 6\times {10}^{-9}}{(0.2{)}^2}\hat{i}=-125\hat{i}N/C.$

$\Rightarrow \overrightarrow{E_B}=\frac{-K{q}_B}{2\sqrt{2}{d}^2}\hat{i}-\frac{K{q}_B}{2\sqrt{2}{d}^2}\hat{j}=-\frac{9\times {10}^9\times 6\times {10}^{-9}}{2\times \sqrt{2}\times (0.2{)}^2}\hat{i}=-\frac{9\times {10}^9\times 6\times {10}^{-9}}{2\times \sqrt{2}\times (0.2{)}^2}\hat{i}-\frac{9\times {10}^9\times 6\times {10}^{-9}}{2\times \sqrt{2}\times (0.2{)}^2}\hat{j}=-\frac{50}{\sqrt{2}}\hat{i}-\frac{50}{\sqrt{2}}\hat{j}$

resultant electric field intensity

$\Rightarrow \overrightarrow{E}=\overrightarrow{E_A}+\overrightarrow{E_B}+\overrightarrow{E_C}=(-125-\frac{50}{\sqrt{2}})\hat{i}+(-50-\frac{50}{\sqrt{2}})\hat{j}=-160.3\hat{i}-85.3\hat{j}$

$\Rightarrow E=181.5N/C$