Question

Positive charges of 9 nC and 4 nC are placed at thee points A and C respectively of a right angled triangle ABC in which $\angle$ b = 90, AB = 3 cm and BC = 2 cm. Find the electric intensity at B.

Answer 1

Let 4C charge which is placed at C has an electric field intensity $-E_{1}\hat{i} at position B

and electric field intensity due to 9C charge which is placed at position A is $-E_{2}\hat{j} at position B

$\Rightarrow E_1=\frac{K{q}_1}{r^2}=\frac{9\times {10}^9\times 4\times {10}^{-9}}{2^2\times {10}^{-4}}=9\times {10}^{4}N/C.$

$\Rightarrow E_2=\frac{K{q}_2}{r^2}=\frac{9\times {10}^9\times 9\times {10}^{-9}}{3^2\times {10}^{-4}}=9\times {10}^{4}N/C.$

resultant $E=-E_1\hat{i}-E_2\hat{j}=-9\times {10}^4\hat{i}-9\times {10}^4\hat{j}=\sqrt{9^2+9^2}\times {10}^4.$

$\Rightarrow 3\sqrt{2}\times {10}^{4}N/C$