Question

Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm. (a) What is the magnitude of the force that each sphere experiences? (b) The spheres are brought into contact and then separated to a distance of 3.59 cm. Determine the magnitude of the force that each sphere now experiences.

Answer 1

(a):
`\rm -5.627* 10^3\ N.`

(b):
`\rm 7.626* 10^2\ N.`

Step-by-step explanation:

Given:

• Charge on one sphere,
  `\rm q_1 = -19.8\ \mu C = -19.8* 10^(-6)\ C.`

• Charge on second sphere,
  `\rm q_2 = +40.7\ \mu C = +40.7* 10^(-6)\ C.`
• Separation between the spheres,
  `\rm r=3.59\ cm = 3.59* 10^(-2)\ m.`

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

`\rm F=k\cdot(q_1q_2)/(r^2)`

where,

k is called the Coulomb's constant, whose value is
`\rm 9* 10^9\ Nm^2/C^2.`

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

`\rm F=k\cdot(q_1q_2)/(r^2)\\=9* 10^9* ((-19.8* 10^(-6))* (+40.7* 10^(-6)))/((3.59* 10^(-2))^2)\\=-5.627* 10^3\ N.`

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres,
`\rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.`

The new charges on both the spheres are equal and given by

`\rm q_1'=q_2'=\frac Q2 = (20.9)/(2)\ \mu C=10.45\ \mu C = 10.45* 10^(-6)\ C.`

The magnitude of the force that each sphere now experiences is given by

`\rm F'=k\cdot (q_1'q_2')/(r^2)'\\=9* 10^9* (10.45* 10^(-6)* 10.45* 10^(-6))/((3.59* 10^(-2))^2)\\=7.626* 10^2\ N.`