91Ó°ÊÓ

• a) Charges of the three spheres,$q_A=-2.00nC,q_B=-4.00nCand{q}_C=+8.00nC.$.
• b)Side length of the triangle where spheres are located at vertices,$d=20.0cm.$.
• c) The steps A and B are connected by wire and then disconnected. B is grounded; B and C are connected and disconnected.

We can get any electrostatic force between two spheres by taking the concept of Coulomb's law. Again using the process of conduction by contact, we can get the individual charges of the spheres, and then using these charges,their force can be found.

Formula:

The magnitude of the electrostatic force between any two particlesis given by

$\left|\stackrel{→}{F_{12}}\right|=\frac{\left|q_1{q}_2\right|}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{d}^2}\left(i\right)$

Using the given data in equation (1), the electrostatic force between spheres A and C is givenby

.$$\begin{gathered} \left|F_{AC}\right|=\frac{\left(9.00×{10}^{9}N{m}^2/C^2\right)\left(-2.00×{10}^{-9}C\right)\left(8.00×{10}^{-9}C\right)}{{\left(0.200m\right)}^2} \\ =3.60×{10}^{-6}N\left(i\right) \end{gathered}$$

Hence, the value of the force is$3.60×{10}^{-6}N$.

After making contact with each other, both A and B each have a charge of

$$\begin{gathered} \frac{q_A+q_B}{2}=\left(\frac{-2.00+\left(-4.00\right)}{2}\right)nC \\ =-3.00nC \end{gathered}$$

When B is grounded, its charge becomes zero.

After making contact with C, which has a charge of +8.00 nC, B and C both acquire a charge of

$$\begin{gathered} \frac{q_B+q_C}{2}=\left[0+\left(-8.00nC\right)\right]/2 \\ =-4.00nC \end{gathered}$$

Finally, we have$Q_A=-3.00nCand{Q}_B=Q_C=-4.00nC.$

Thus, the electrostatic force between the spheres A and C using equation (i) is givenby

localid="1663160030491" $$\begin{gathered} \left|F_{AC}\right|=\frac{\left(9.00×{10}^{9}N{m}^2/C^2\right)\left(-3.00×{10}^{-9}C\right)\left(-4.00×{10}^{-9}C\right)}{{\left(0.200m\right)}^2} \\ =2.70×{10}^{-6}N \end{gathered}$$

Hence, the value of the force is$2.70×{10}^{-6}N$.

The electrostatic force between the spheres A and C using equation (i) is givenby

$$\begin{gathered} \left|F_{ac}\right|=\frac{\left(9.00×{10}^{9}N{m}^2/C^2\right)\left(-4.00×{10}^{-9}C\right)\left(-4.00×{10}^{-9}C\right)}{{\left(0.200m\right)}^2} \\ =3.60×{10}^{-6}N \end{gathered}$$

Hence, the value of the force is$3.60×{10}^{-6}N$ .