91Ó°ÊÓ

The value of the electrostatic force of attraction,$F_1=0.108N$.

The separation between the two identical spheres, $r=50.0cm\left(\frac{(1 m)}{100 cm}\right)=0.5 m$

After being connected through a thin wire and then removed, the force of repulsion, $F_2=0.0360N$

According to Coulomb's Law of electrostatic attraction or repulsion within particles, the force acting on them is given as being directly proportional to the product of the charges on the particles and being inversely proportional to the separation between them. Using this concept, we can find out the force acting on them.

Formula:

The magnitude of the electrostatic force between any two particles,

$F_1=k\left|q_1\right|\left|q_2\right|/r^2$ (1)

Since opposite charges attract the initial charge configurations must be of opposite signs. Similarly, since like charges repel, the final charge configurations must carry the same sign.

We assume that the spheres are far apart. Then the charge distribution on each of them is spherically symmetric and Coulomb’s law can be used. Let, q₁ and q₂ be the original charges of the spheres. We choose the coordinate system so the force on q₂ is positive if it is repelled by q₁. Then the force on q₂ in the first case of attraction is given as:

$$\begin{gathered} F_a=-k\left|q_1\right|\left|q_2\right|/r^2........................................\left(2\right) \\ (negativesignindicatestheattractiveforce) \end{gathered}$$

After the wire is connected, the spheres, being identical, acquire the same charge. Since the charge is conserved, the total charge is the same as it was originally. This means the charge on each sphere is localid="1662705665825" $\frac{|q_1+q_2|}{2}$. The force is now repulsive and is given by using equation (1) as follows:

localid="1662708778201" $$\begin{gathered} F_b=k\frac{\left|(q_1+q_2)/2\right|\left|(q_1+q_2)/2\right|}{r^2} \\ =k\frac{{\left|(q_1+q_2)\right|}^2}{4r^2}......................................\left(3\right) \end{gathered}$$

Using equation (2) and the given values, the product of charges is found to be

$$\begin{gathered} q_{1}q2=-\frac{r^2{F}_a}{k} \\ =-\frac{{(0.500m)}^2(0.018N)}{9×{10}^{9}N.m^2/C^2} \\ =-3.00×{10}^{-12}{C}^2 \end{gathered}$$

Now, using the given values in equation (3), the sum of the charges is given as:

$q_1+q_2=2r\sqrt{\frac{F_b}{k}}$

$$\begin{gathered} =2(0.5m)\sqrt{\frac{0.0360N}{9×{10}^{9}N.m^2/C^2}} \\ =2.00×{10}^{-6}C........................\left(5\right) \end{gathered}$$

where we have taken the positive root (which amounts to assuming q1 + q2 $â‰\yen$0).

Using equation (3), the value of the charge of the second sphere is given as:

$q_2=\frac{-(3.00×{10}^{-12}{C}^2)}{q_1}$

Substituting the above value in equation (5), we can get the value of the first charge aslocalid="1662708533177" $$\begin{gathered} q_1-\frac{(3.00×{10}^{-12}{C}^2)}{q_1}=2.00×{10}^{-6}C \\ {q}_1^2-(2.00×{10}^{-6}C)q_1-3.00×{10}^{-12}{C}^2=0 \\ {q}_1=\frac{2.00×{10}^{-6}CÂ\pm √\left(\right(2.00×{10}^{-6}C{)}^2-4(-3.00×{10}^{-12}{C}^2)}{2} \\ {q}_1=3.00×{10}^{-6}C \end{gathered}$$

Substituting the above value in the value of the second charge, we get the other charge as:

$q_2=-1.00×{10}^{-6}C$

Hence, the value of the negative charge is $-1.00×{10}^{-6}C$

From the calculations of the part (2), we can get the positive charge of the other sphere is 3.00$×{10}^{-6}C$