Solve for the net intensity at point A.
q1 9 m A 2 m q2
I
. _______________________________O__________________________.
-5 µC + 3 µC
Answer
Net electric field intensity at point a can be given as
E=kq1r12+kq2r22E=k(q1r12+q2r22)E=\frac{kq_1}{r_1^2}+\frac{kq_2}{r_2^2}\\E=k(\frac{q_1}{r_1^2}+\frac{q_2}{r_2^2})E=r12kq1+r22kq2E=k(r12q1+r22q2)
By putting value
E=(9×109)(5×10−692+3×10−622)E=(9\times10^9)(\frac{5\times10^{-6}}{9^2}+\frac{3\times10^{-6}}{2^2})E=(9×109)(925×10−6+223×10−6)
E=7.3×10−3N/cE=7.3\times10^{-3}N/cE=7.3×10−3N/c
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