Question

Three charges of respective values $-\sqrt{2}\mu C,2\sqrt{2}\mu C$ and $-\sqrt{2}\mu C$ are arranged along a straight line as shown in the figure. Calculate the total electric field intensity due to all three charges at the point P.

Answer 1

Electric field at

(i) Point P due to charge at A or C is

$E_{PA}=E_{PC}=9\times {10}^9\times \frac{\sqrt{2}\times {10}^{-6}}{(\sqrt{2}{)}^2}$

$=\frac{9\sqrt{2}}{2}\times {10}^{3}N{C}^{-1}$

(ii) Point P due to charge at B is

$E_{PB}=9\times {10}^9\times \frac{2\sqrt{2}\times {10}^{-6}}{(1{)}^2}$

$=18\sqrt{2}\times {10}^{3}N{C}^{-1}$

Horizontal component of net electric field at point P is

$E_X=E_{PC}\cos \theta -E_{PA}\cos \theta =0$

Vertical component of net electric field at point P is

$E_Y=E_{PB}-E_{PA}\sin \theta$

$E_Y=[18\sqrt{2}-\frac{9\sqrt{2}}{2}\times \frac{1}{\sqrt{2}}-\frac{9\sqrt{2}}{2}\times \frac{1}{\sqrt{2}}]\times {10}^{3}N{C}^{-1}$

$E_Y=16.45\times {10}^{3}N{C}^{-1}$

So, resultant electric field at point P is

$E=\sqrt{E_X^2+E_Y^2}=\sqrt{0+(16.45\times {10}^3{)}^2}$

$E=16250N{C}^{-1}$