Grade 11Electrostatics

plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz solve question no 50

Question image for plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz solve q

YASH AHUJA

8 Years agoGrade 11

1 Answer

Eshan

8 Years ago

Electric field at D due to charge at point A is towards C of magnitude= $E_1=\dfrac{1}{4\pi\epsilon_0}\dfrac{10^{-6}}{1^2}$
Electric field at D due to charge at point C is towards A of magnitude= $E_2=\dfrac{1}{4\pi\epsilon_0}\dfrac{2\times 10^{-6}}{1^2}$
Hence the two electric fields will add up to give $E'=\dfrac{1}{4\pi\epsilon_0}\dfrac{1\times 10^{-6}}{1^2}$
Electric fied at D due to charge at point B is perndicular to line AC of magnitude $E_3=\dfrac{1}{4\pi\epsilon_0}\dfrac{5\times 10^{-6}}{(\sqrt{3})^2}$
Hence the magntiude of net force is $\sqrt{E'^2+E_3^2}=1.75\times 10^4NC^{-1}$