Question

Question: Of the chargeQon a tiny sphere, a fractionis to be transferred to a second, nearby sphere. The spheres can be treated as particles. (a) What value ofmaximizes the magnitudeFof the electrostatic force between the two spheres? What are the (b) smaller and (c) larger values ofthat putFat half the maximum magnitude?

Short answer

• a)The force between the two spheres will be maximum when$\alpha =0.5$ .
• b)Smaller value of$\alpha$for which the force has half the maximum magnitude is${\alpha }_1=0.15.$.
• c) Larger value of $\alpha$for which the force has half the maximum magnitude is${\alpha }_2=0.85.$ .

Step-by-step solution

Given

Charge on the tiny sphere is .

Step 2: Coulomb’s law

According to Coulomb’s law, the electrostatic force F between the two point charges is directly proportional to the product of charges qand Q inversely proportional to the square of the distance rbetween them.

(a) Find out what value of a maximizes the magnitude F of the electrostatic force between the two spheres

The two charges are$q=\alpha Q$(where$\alpha$is a pure number presumably less than 1 and greater than zero) and$Q-q=(1-\alpha )Q$ .

using coulomb’s law,

$$\begin{gathered} F=\frac{\left|\alpha Q\right|\left|1-\alpha \right|\left|Q\right|}{4\pi {\epsilon }_0{d}^2} \\ =\frac{Q^2\alpha \left|1-\alpha \right|}{4\pi {\epsilon }_0{d}^2} \end{gathered}$$

The graph below, ofversus$\alpha$ , has been scaled so that the maximum value of force is 1. The actual value of force is-

$F_{max}=\frac{Q^2}{16\pi {\epsilon }_0{d}^2}$

It is clear that $\alpha =1/2$ gives the maximum value of F.

(b) Calculate the smaller value of  that put F at half the maximum magnitude

Without the gridlines it is difficult to calculate the half points from the graph, though some calculators are capable of doing that. But we can easily calculate the half points algebraically using the quadratic formula. The results are

$$\begin{gathered} a_1=\left(1/2\right)\left(1-\frac{1}{\sqrt{2}}\right)\approx 0.15and{\alpha }_2=\left(1/2\right)\left(1+\frac{1}{\sqrt{2}}\right)\approx 0.85 \\ Thus,thesmallervalueofais{\alpha }_1=0.15. \end{gathered}$$.

(c) Calculate the larger value of  that put F at half the maximum magnitude

${\alpha }_1=\left(1/2\right)\left(1-\frac{1}{\sqrt{2}}\right)\approx 0.15and{\alpha }_2=\left(1/2\right)\left(1+\frac{1}{\sqrt{2}}\right)\approx 0.85$

Thus, Larger value of $\alpha is{\alpha }_2=0.85$