Question

Consider two point charges q1 = q2 = 10.0 C are located on the x – axis, as
shown in the following figure. If a = 0.10 m, find the electric field at point P in unit-vector
notation.

Answer 1

$E = E_{1}j +E_{2}Cos 45^{0}j+E_{2}Sin45^{0}(-i)$

$E =(kq_{1}/a^{2}+kq_{2}/2a^{2})j+(kq_{2}/2a^{2} \times 1/\sqrt{2} (-i))$

$=(\frac{9\times 10^{9}\times 10\times 10^{-6}}{0.01}+ \frac{9 \times 10^{9} \times 10\times 10^{-6}}{2(0.01)\sqrt{2}})j$ $+ (\frac{9\times 10^{9}\times 10\times 10^{-6}}{2(0.01)\sqrt{2}})(-i)$

$(9\times 10^{6} +3.18\times10^{6})j+(3.18 \times 10^{6})(-i)$

$=(3.18(-i)+12.18j)\times 10^{6}N/C$