91Ó°ÊÓ

a) The combined charge of the two small positively charged spheres,$Q=5.0×{10}^{−5}\text{C}$

b) Electrostatic force on each sphere due to repulsion,$F=1.0\text{N}$

c) Separation between the charges,$r=2.0\text{m}$

We know that two like charges repel while unlike charges attract each other, giving rise to an electrostatic force on each other due to charged bodies. Thus, here, with two equally like charges, we see that both spheres impose a repulsion force on each other. Considering one sphere, we can get the force given by the other on its body.

The electrostatic force of attraction or repulsion,

$F=\frac{k{q}_1{q}_2}{r^2}$ (i)

where $k=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}=9.0×{10}^9{\text{N-m}}^2/{\text{C}}^2$, is the electrostatic constant, is the charge of first sphere,$q_2$ is the charge on the second sphere, is the separation between the spheres.

Let the two positive charges be$q_1$and$q_2$.

Then, we are given the combined charge value as:

$\begin{array}{c}{q}_1+q_2=Q\\ =5.0×{10}^{−5}\text{ C â¶Ä‰}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(\text{i}\right)\end{array}$

Again, the electrostatic force due to repulsion on each other can be given using the given data and equation (i) as follows:

$\begin{array}{l}1.0\text{±·â€‰}=\frac{(9.00×{10}^9{\text{ N-³¾}}^{\text{2}}{\text{/C}}^{\text{2}})q_1{q}_2}{{\left(2.0\text{″¾}\right)}^2}\\ q_1{q}_2=0.44×{10}^{−9}{\text{C}}^2\text{ â¶Ä‰â¶Ä‰}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(\text{ii}\right)\end{array}$

Now, using equation (i) and (ii), we can get the difference in their charged values as follows: (consider $q_1$as the larger charge)

$\begin{array}{c}{(q_1−q_2)}^2={(q_1+q_2)}^2−4\left(q_1{q}_2\right)\\ ={(5×{10}^{−5}\text{C})}^2−4(0.44×{10}^{−9}{\text{C}}^2)\\ =2.5×{10}^{−9}{\text{C}}^2−1.76×{10}^{−9}{\text{C}}^2\\ =0.74×{10}^{−9}{\text{C}}^2\end{array}$

On solving further,

$\begin{array}{c}(q_1−q_2)=\sqrt{0.74×{10}^{−9}{\text{C}}^2}\\ =2.72×{10}^{−5}\text{°ä â¶Ä‰}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(\text{iii}\right)\end{array}$

Now, adding equations (I) and (III), we can get the charge of first sphere as follows:

$\begin{array}{c}(q_1+q_2)+(q_1−q_2)=5×{10}^{−5}\text{C}+2.72×{10}^{−5}\text{C}\\ 2q_1=7.72×{10}^{−5}\text{C}\\ q_1=\frac{7.72×{10}^{−5}\text{C}}{2}\\ q_1=3.86×{10}^{−5}\text{C}\end{array}$

Now, using the above value in equation (I), we get the charge of the smaller sphere as follows:

$\begin{array}{c}{q}_2=Q−q_1\\ =5×{10}^{−5}\text{C}−3.86×{10}^{−5}\text{C}\\ =1.14×{10}^{−5}\text{C}\end{array}$

Thus, the sphere carrying less charge,has $1.14×{10}^{−5}\text{C}$charge.