Question

three electric charges q0,q1 and q2 are distances r0,r1 and r2 with respect to same origin. what is the force on the charge q0 in the field of charges q1 and q2?

Answer 1

Answer: The total force on q in the field of charged q1 and q2 is given by:

Explanation:

Answer 2

Given :

Three electric charges as :

$q_{0} ,q_{1}$ and $q_{2}$

Distances of the three charges with respect to the origin :

$r_{0},r_{1}$ and $r_{2}$ respectively

To Find :

The force on the charge $q_{0}$ in the field of charges $q_{1}$ and $q_{2}$ = ?

Solution :

∴The  total electrostatic force on the charge $q_{0}$ will be due to the charges $q_{1}$ and $q_{2}$ .

Now the distance between $q_{0}$ and $q_{1}$ is :

$r_{1} - r_{0}$

And the distance between $q_{0}$ and $q_{2}$ is :

$r_{2} - r_{0}$

Now according to Superposition theorem the total force on $q_{0}$ due to $q_{1}$ and $q_{2}$ will be equal to the sum of the forces due to $q_{1}$ and $q_{2}$ individually on $q_{0}$ .

So the total force will be :

$F_{T} = F_{q_{1}}+F_{q_{2$

    $= \frac{K q_{0} q_{1}}{(r_{1}-r_{0})^{2} }$ +$\frac{K q_{0} q_{2}}{(r_{2}-r_{0})^{2} }$

$= Kq_{0}\times [ \frac{q_{1}}{(r_{1} -r_{2})^{2} } +\frac{q_{2}}{(r_{2}-r_{0})^{2} } ]$

So finally the total electrostatic force acting on  $q_{0}$   is $= Kq_{0}\times [ \frac{q_{1}}{(r_{1} -r_{2})^{2} } +\frac{q_{2}}{(r_{2}-r_{0})^{2} } ]$  .