Question

In a triangle APC, AC =AB+BC, AB=BC=2cm.AP=4cm,PC=4cm, At point B BP is perpendicular to AC. Charges q1,q2,and q3 are placed at A, B, C respectively. And q1 =q2=-q3=2micro C. Determine the magnitude and the direction of the electric field at point P.

Answer 1

As per the question,

In the tirangle ABC,

AC= AB+BC,

So, $F_{AC}=F_{AB}+F_{BC}=\dfrac{q_1q_2}{4\pi \epsilon_o R^2}+\dfrac{q_2q_3}{4\pi \epsilon_o R^2}=\dfrac{9\times 4\times10^{-3}}{4\times 10^{-4}}+\dfrac{9\times 4\times10^{-3}}{4\times 10^{-4}}N$

$F_{AC}=180N$

Given that AC= AB+BC=2+2=4cm

Hence 180=$\sqrt{{\dfrac{Qq_1}{4\pi \epsilon r^2}}^2+{\dfrac{Qq_1}{4\pi \epsilon r^2}}^2+2{\dfrac{Qq_1}{4\pi \epsilon r^2}}{\dfrac{Qq_1}{4\pi \epsilon r^2}}\cos60}={\dfrac{Qq_1}{4\pi \epsilon r^2}}\sqrt{3}$

$Q=\dfrac{180\times 16\times 10^{-4}}{9\times 10^{9}\times \sqrt{3}\times 2\times 10^{-6}}=92.37\times 10^{-6}C$