Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108 N when separated by 50.0 cm, center-to-center. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.360 N. What were the initial charges on the spheres?

Answer 1

Answer:

The initial charges of the spheres were q₁=6.7712×10⁻⁶C and q₂=-4.4350×10⁻⁶C.

Explanation:

As the spheres attract each other, the charges of the spheres are opposite.

The atracction force is given by:

F=$-\frac{Kq_{1}q_{2}}{r^{2}}$

where:

K: Coulomb constant

q₁: charge of sphere 1

q₂: charge of sphere 2

r: distance between both charges

The electrostatic atraction force is 0.108 N so:

0.108N=-8.99×10⁹$\frac{N*m^2}{C^2}$ $\frac{q_{1}q_{2}}{(0.5m)^2}$

q₁·q₂=$-\frac{0.108N*(0.5m)^2}{8.99*10^9\frac{N*m^2}{C^2}}$

q₁·q₂=-3.003×10⁻¹² C²

When the wire is connected the charges are equally distributed as the spheres are identical. Hence, the final charge is of each sphere is $\frac{q_1q_2}{2}$

The repel force is 0.360 N and it is given by:

F=$\frac{K(\frac{q_1+q_2}{2})(\frac{q_1+q_2}{2})}{r^{2}}$

F=$\frac{K(q_1+q_2)^2}{4r^{2}}$

Then, we get a secong equation:

(q₁+q₂)²=$\frac{0.360N*4*(0.5m)^2}{8.99*10^9\frac{N*m^2}{C^2}}$

(q₁+q₂)=√4.004×10⁻¹¹ C²

q₁+q₂=6.3277×10⁻⁶ C

We solve the equation system:

$\left \{ {{q_1=\frac{-3.003*10^{-12}C^2}{q_2} } \atop {q_1+q_2=6.3277*10^{-6}C}} \right.$

We replace q₁ in the second equation:

$\frac{-3.003*10^{-12}C^2}{q_2} +q_2=6.3277*10^{-6}C$

$-3.003*10^{-12}C^2+(q_2)^2=6.3277*10^{-6}C*q_2$

$(q_2)^2-6.3277*10^{-6}C*q_2-3.003*10^{-12}C^2=0$

The solutions are:

q₁=6.7712×10⁻⁶C

q₂=-4.4350×10⁻⁶C