# Two charges q1=4*10^-11 C and q2=-6*10^-11 C are in vacuum seperated between each other by a distance of 0.2 cm. What is the Vector Force that the first charge q1 exerts on q2. answer. -54*10^-11N...

Two charges q1=4*10^-11 C and q2=-6*10^-11 C are in vacuum seperated between each other by a distance of 0.2 cm. What is the Vector Force that the first charge q1 exerts on q2.

answer. -54*10^-11N * r012

### 1 Answer | Add Yours

The mathematical definition of electrostatic field is derived from Coulomb’s law which defines the vector force between two point charges kept in vacuum. Given point charges `q_1, q_2` located by the vectors `R_1` and `R_2` respectively, the vector force acting on `q_2` due to `q_1` is defined by Coulomb’s law as:

`F_(12)=(q_1*q_2)/(4*pi*epsilon_0*(|R_2-R_1|^2))*r_0(12)`

where `r_0(12)` is a unit vector pointing from `q_1` to `q_2` , |R_2-R_1|, the distance between the charges and `epsilon_0` is the free space permittivity (`=8.854*10^(-12)F/m^2` )

Put the values of the given problem to get the force vector as:

`F_(12)=(4*10^(-11)* (-6)*10^(-11))/(4*pi*8.854*10^(-12)*(2*10^(-3))^2)*r_0(12)`

`=-5.4*10^-6 N * r_0(12)`

**Sources:**

### Hide Replies ▲

Did you calculate everything right, im getting a different result

-24*10^-22/222.41*10^-18 = 1*10^-5