Two identical conducting spheres carry charges of +5.0 $\mu$C and -1.0 $\mu$C, respectively. The centers of the spheres are initially separated by a distance L. The two spheres are brought together so that they are in contact. The spheres are then returned to their original separation L. What is the ratio of the magnitude of the electric force on either sphere after the spheres are touched to that before they were touched?

Question

Quizplus · Accepted Answer

Before the spheres are touched, the magnitude of the electric force on each sphere is given by:

$F_1 = F_2 = \frac{kq^2}{L^2}$

After the spheres are touched, they will have the same charge of $Q=5-1=4\,\mu C$. The total charge is conserved, so each sphere will have a final charge of $2\,\mu C$.

The electric force between the spheres after they are touched is given by:

$F' = \frac{kq^2}{4L^2}$

The ratio of the electric forces before and after touching is:

$\frac{F'}{F_1} = \frac{\frac{kq^2}{4L^2}}{\frac{kq^2}{L^2}} = \frac{1}{4}$

Similarly, the ratio of the electric forces after touching and before touching is:

$\frac{F_2}{F'} = \frac{\frac{kq^2}{L^2}}{\frac{kq^2}{4L^2}} = \frac{4}{1}$

Multiplying these ratios, we get:

$\frac{F_2}{F_1} = \frac{F_2}{F'} \cdot \frac{F'}{F_1} = \frac{4}{1} \cdot \frac{1}{4} = \boxed{	extbf{(B) } \frac{4}{5}}$

Two Identical Conducting Spheres Carry Charges of +5 μ\muμ C and -10

Two Identical Conducting Spheres Carry Charges of +5 $\mu$ C and -10