Question

Two very small spheres are initially neutral and separated by a distance of 0.50m. Suppose that 3.0 x 1013 electrons are removed from one sphere and placed on the other.
A.What is the magnitude of the electrostatic force that acts on each sphere? (0.83 N)
B.Is the force attractive or repulsive? Why? (attractive)

Answer 1

Answer:

A) 0.83 N B) It is attractive.

Explanation:

A)

• If we remove 3.0*10¹³ electrons from a neutral sphere, this means that the sphere will acquire a positive charge, equal to the total charge removed.
• As each electron carries a charge equal to -e= -1.6*10⁻¹⁹ C, the total charge on the first sphere is as follows:

       $Q_{1} = 1.6e-19 C * 3.0e13 electrons = 4.8e-6 C$

• The second sphere, will acquire an exactly equal charge, but of opposite sign, as there will be a negative net charge on it:

       $Q_{2} =( -1.6e-19 C) * 3.0e13 electrons) = -4.8e-6 C$

• Assuming both spheres can be treated as point charges, the force between them, must obey Coulomb's Law.
• Applying Coulomb's Law to both charged spheres, we can find the magnitude of the force between them as follows:

       $F =\frac{k*Q_{1}*Q_{2}}{r^{2} } = \frac{9e9N*m2/C2*(4.8e-6C)^{2}}{(0.5m)^{2} } = 0.83 N$

• The magnitude of the electrostatic force that acts on each sphere is 0.83N.

B)

• The force is attractive, because after removing negative charge from one sphere, this acquires a positive charge, and as the charge removed from the first sphere is placed on the other, this acquires a negative charge.
• As opposite charges attract each other, the force between both spheres is attractive.