The sum of two point charges is +8mC. When they are 0.3m apart each experiences a force of 150 N. Find the charges given that the force is:

Question

anonymous · Answer

$$F= \ \frac{ kq _{1} q _{2}}{ r ^{2}} = \frac{ 9x10^{9} q _{1} q _{2}}{ 0.3 ^{2}} = 150$$

festinger · Answer

$$ q_{1}+q_{2}=+8mC$$
So$$q_{1}*q_{2} = q_{1}*(8mC-q_{1})$$

You will have to solve a quadratic equation for this.

anonymous · Answer

can you explain how you got to the second equation?

anonymous · Answer

it should e 0.03m

festinger · Answer

$$q_{1}+q{2}=+8mC$$
Adding q1 on both sides...
$$q_{1} + q{2} - q_{1} = +8mC - q_{1}$$
$$q_{2} = 8mC - q_{1}$$

Sub this into q1*q2 to get the 2nd expression.

anonymous · Answer

Hey thanks I was actually converting cm to m incorrectly, if i hadn't i would of had the correct answer. 


by my equation i would be able to get $$q _{1}*q _{2} = 15 and q _{1} + q _{2} = 8 and then use a quadratic \to solve that. q ^{2} + 8q = 15 = 0 then i get q = 5 q = 3 $$