Answered By Jules E

To find the resulting charge on the spheres after they touch and are moved apart, simply add the charges and divide by the number of spheres (in this case, 2).

    $\frac{\text{(4.50uC)+(-2.40uC)}}{2}=1.05\mu C$(4.50uC)+(-2.40uC)2 =1.05μC.

I'm not sure if you are solving for the electrostatic force since that part of the question has been cut off. If you are, you use the following formula:

 $\left|F_e\right|=\frac{kq_1q_2}{^{r^2}}$|F_e|=kq₁q₂^r2‍  where k is the Coulomb's Law constant of 8.99x10⁹ Nm²/C².

Remember to convert the charge from  $\mu C$μC to C when solving for F_e. Also convert cm to m.

   $\left|\text{Fe}\right|=\frac{\left(8.99\times10^9\frac{Nm^2}{C^2}\right)\left(1.05\times10^{-6}C\right)\left(1.05\times10^{-6}C\right)}{\left(2.5\times10^{-2}m\right)^2}$|Fe|=(8.99×10⁹Nm²C² )(1.05×10⁻⁶C)(1.05×10⁻⁶C)(2.5×10⁻²m)²  

 $\left|F_e\right|=15.9N$|F_e|=15.9N 

Therefore the electrostatic force is 15.9 N  

Answered By Clifton P

Because the spheres have identical surface areas we can say that momentary contact will spread the charge equally between them. This means each sphere will have

q_ave=  $\frac{q1+q2}{2}=\frac{4.5\mu C+\left(-2.4\mu C\right)}{2}=1.05\mu C$q1+q22 =4.5μC+(−2.4μC)2 =1.05μC 

Now that we have the charge we can find the force between them.

 $F=\frac{kqq}{r^2}=\frac{\left(8.99x10^9\right)\left(1.05x10^{-6}C\right)\left(1.05x10^{-6}C\right)}{\left(0.025m\right)^2}$F=kqqr² =(8.99x10⁹)(1.05x10⁻⁶C)(1.05x10⁻⁶C)(0.025m)²  

 $F=15.85N=15.9N$F=15.85N=15.9N 

Because the spheres have the same charge, we know this is a repulsive force, pushing them away from each other.