Question

Question: In Fig. 21-41, three identical conducting spheres form an equilateral triangle of side length d=20.0 cm. The sphere radii are much smaller thand, and the sphere charges are, $q_A=-2.00nC,q_B=-4.00nC,and{q}_c=+8.00nC.$

(a) What is the magnitude of the electrostatic force between spheresAandC?

The following steps are then taken:AandBare connected by a thin wire andthen disconnected;Bis grounded by the wire, and the wire is then removed;BandCare connected by the wire and then disconnected. What now are the magnitudes of the electrostatic force (b) between spheresAandCand (c) between spheresBandC?

Short answer

• a)The magnitude of electrostatic force between spheres A and C before the stepsare taken is $3.60\times {10}^{-6}N$.
• b)The magnitude of the electrostatic force between spheres A and C after the steps are taken is $2.70\times {10}^{-6}N$.
• c) The magnitude of the electrostatic force between spheres B and C after the steps are taken is$3.60\times {10}^{-6}N$ .

Step-by-step solution

Step 1: Stating given data

• 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.

Understanding concept of Coulomb’s law

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

$\left|\overrightarrow{F_{12}}\right|=\frac{\left|q_1{q}_2\right|}{4\pi {\epsilon }_0{d}^2}\left(i\right)$

a) Calculation of force between A and C before steps were taken

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

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

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

b) Calculation of force between A and C after steps were taken

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

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

c) Calculation of force between B and C after steps were taken

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

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

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