Question

. Two charges and B of magnitudes +8.0*10^6C each,are kept at points (2,1) and (2,7) respectively in a Cartesian plane measured in meters .if the charges are in vacuum , calculate the force of A on B*

Answer 1

Coulomb’s law describes the electrostatic force (or electric force) between two charged particles. If the particles have charges $q_1$ and $q_2,$ are separated by distance $r,$ and are at rest (or moving only slowly) relative to each other, then the magnitude of the force acting on each due to the other is given by

$$\bar{F}={1\over 4\pi \varepsilon_0\varepsilon}\cdot{q_1q_2\over r^3}\bar{r}$$

Given

$q_1=q_2=+8.0\times10^{-6} C,$

$(x_A, y_A)=(2, 1), (x_B, y_B)=(2, 7)$

$\varepsilon=1$

$\dfrac{1}{4\pi \varepsilon_0}=9\times10^9\ \dfrac{N\cdot m^2}{C^2}$

Then

$$\bar{r}=(x_B-x_A, y_B-y_A)=(2-2, 7-1)=(0,6)$$

$$t=|\bar{r}|=\sqrt{(0)^2+(6)^2}=6$$

$$|\bar{F}|=9\times10^9\ {N\cdot m^2\over C^2}\cdot{(8.0\times10^{-6} C)^2\over (6 m)^2}=1.6\times10^{-2} \ C$$

The magnitude of the electrostatic force of repulsion is equal to $1.6\times10^{-2} \ C.$