Question

A charge of -2.90 nC is placed at the origin of an xy-coordinate system, and a charge of 2.35 nC is placed on the y axis at y = 3.80 cm .

Part A: If a third charge, of 5.00 nC , is now placed at the point x = 3.45 cm , y = 3.80 cm find the x and y components of the total force exerted on this charge by the other two charges.

Part B: Find the magnitude of this force.

Part C: Find the direction of this force.

Please help I cant figure this one out i have used the same formula and keep getting the same numbers so I dont kknwo what I am doing wrong,.

Answer 1

A)

  Total force is
   $\vec{F}=\frac{Q}{4\pi \epsilon_0}\left [ \frac{q_1(\vec{r}-\vec{r}_1)}{|\vec{r}-\vec{r}_1|^3}+ \frac{q_2(\vec{r}-\vec{r}_2)}{|\vec{r}-\vec{r}_2|^3}\right ]$
$\Rightarrow \vec{F}=5\times 10^{-9}\times 9\times 10^{9}\times \left [ \frac{(-2.9\times 10^{-9})(3.45\hat{x}+3.80\hat{y}-\vec{0})\times 10^{-2}}{|3.45\hat{x}+3.8\hat{y}-\vec{0}|^3\times 10^{-6}}\\ +\frac{(2.35\times 10^{-9})(3.45\hat{x}+3.80\hat{y}-3.8\hat{y})\times 10^{-2}}{|3.45\hat{x}+3.8\hat{y}-|^3\times 10^{-6}}\right ]$ $\Rightarrow \vec{F}=45\times 10^{-5}\times \left [ -\frac{2.9\times (3.45\hat{x}+3.80\hat{y})}{(3.45^2+3.8^2)^{3/2}}\\ +\frac{2.35(3.45\hat{x})}{|3.45\hat{x}|^3}\right ]$
$\Rightarrow \vec{F}=45\times 10^{-7}\times \left [ -7.4\hat{x}-8.15\hat{y}\\ +19.74\hat{x})\right ]$
$\Rightarrow \vec{F}=45\times 10^{-7}\times \left [ 12.34\hat{x}-8.15\hat{y}\right ]$
So,
   $\Rightarrow {F}_x=5.55\times 10^{-5}~N,~~~F_y=-3.667\times 10^{-5}~N$
B)
   $\Rightarrow {F}=\sqrt{5.55^2+3.667^2}\times 10^{-5}~N=6.65\times 10^{-5}~N$