Question

Point charge q1 = -5.00 nC is at the origin and point charge q2 = +3.00 nC is on the x-axis at x = 3.00 cm. Point P is on the y-axis at y = 4.00 cm. (a) Calculate the electric fields E S 1 and E S 2 at point P due to the charges q1 and q2 . Express your results in terms of unit vectors (see Example 21.6). (b) Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form.

Answer 1

Q1=-5nC, Q2=3nC, r1=(0; 0; 0), r2=(0.03m; 0; 0), r=(0; 0.04m; 0)

E1(r)=$\frac{Q_{1} }{4\pi \epsilon_{0} } \frac{\vec{r}-\vec{r_{1} }}{|\vec{r}-\vec{r_{1} }|^{3} }=\frac{-5nC}{4\pi 8.85\times 10^{-12}F/m } \frac{(0; 0.04 m; 0)}{6.4\times 10^{-5}m} = (0; -16851.65; 0)\frac{N}{C}$

E2(r)=$\frac{Q_{2} }{4\pi \epsilon_{0} } \frac{\vec{r}-\vec{r_{2} }}{|\vec{r}-\vec{r_{2} }|^{3} }=\frac{3nC}{4\pi 8.85\times 10^{-12}F/m } \frac{(-0.03m; 0.04 m; 0)}{1.25\times 10^{-4}m} = (-6471.03; 8628.04; 0)\frac{N}{C}$

E(r)=E1(r)+E2(r)=(-6471.03; -8223.61; 0)N/C