# 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.

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

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: