Two identical conducting spheres, fixed in place, attract each other with a forc

Question

Two identical conducting spheres, fixed in place, attract each other with a force of 0.080 N when their center to center separation is 54.00 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres have a net positive charge and repel each other with an electrostatic force of 0.035 N. What was the initial negative charge on one of the spheres, and what was the initial positive charge on the other?(Hint: Use charge conservation and solve for one of the initial charges. You will end up with a quadratic equation. The solutions give you the positive and negative charges.)

Explanation / Answer

oulomb's Law:


ec{F}_{12} = rac{kq_{1}q_{2}}{r^2}hat{r}_{12}


Can be rewritten with the scalar component F_{12}


ec{F}_{12} = F_{12}hat{r}_{21}


Also it is known that q1 and q2 are unlike-sign to account for the initial attraction.

Therefore, letting F_{12} = 0.108 N equiv +.

and F'_{12} = 0.0360 N equiv +. We have:


-{F}_{12} = rac{k_{e}q_{1}q_{2}}{r^2}


{F'}_{12} = rac{k_{e}{q'}_{1}{q'}_{2}}{r^2}


Where {q'}_{1} and {q'}_{2} represent the charges after the wire is removed. And so due to Conservation of Charge.


{q'}_{1} = {q'}_{2} = left(rac{{q}_{1} + {q}_{2}}{2} ight)


And therefore,


{F'}_{12} = rac{k_{e}(q_{1}+q_{2})^2}{4{r^2}}


In solving the above for (q_{1}+q_{2}), the problem that arises is whether to choose the positive or negative root to solve for q1 and q2. And justifying that choice.


q_{1}+q_{2} = pmsqrt{rac{4{r^2}{{F'}_{12}}}{{k}_{e}}}

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