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katrin [286]
3 years ago
12

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.
Physics
1 answer:
romanna [79]3 years ago
7 0

Answer:

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

(b):  \rm 7.626\times 10^2\ N.

Explanation:

<u>Given:</u>

  • Charge on one sphere, \rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.
  • Charge on second sphere, \rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.
  • Separation between the spheres, \rm r=3.59\ cm = 3.59\times 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\dfrac{q_1q_2}{r^2}

where,

k is called the Coulomb's constant, whose value is \rm 9\times 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\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 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'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.

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

\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.

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