Question

Two tiny conducting spheres are identical and carry charges of - 20.0 µC and +50.0 µC. They are separated by a distance of 2.50 cm. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? (b) The spheres are brought into contact and then separated to a distance of 2.50 cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive of repulsive.

Answer 1

Answer:

a

The force experience by the two spheres is $F_1 = 1.44*10^4 N$

This force is attractive cause the charge are unlike charges

b

The force experienced by the two spheres is  $F_2 = 3.24*10^{3} \ N$

The force is repulsive because the two charges are like charges

Explanation:

From the question we are told that

   The charge on the first sphere is  $q_1 = -20.0 \mu C = 20 *10^{-6} C$

    The charge on the second sphere is  $q_2 = 50 \mu C = 50*10^{-6} C$

      The distance of separation is $d = 2.50 \ cm = \frac{2.50}{100} = 0.025 \ m$

       

The electrostatic force experienced by the two spheres is mathematically represented as

                  $F_1 = \frac{k q_1 q_2}{d^2}$

Now k is the coulomb’s constant with a value of  $k = 9*10^9 N \cdot m^2 /C^2$

   So

              $F_1 = \frac{9*10^9 * 20 *10^{-6} * 50*10^{-6}}{0.025}$

              $F_1 = 1.44*10^4 N$

When the sphere are brought together the charge on each sphere would be the average of the total charge and this can be mathematically evaluated as

            $q = \frac{q_1 + q_2 }{2}$

            $q = \frac{(-20 + 50)*10^{-6} }{2}$

            $q = 15 \mu C$

So when they are seperated the electrostatic force experienced is  

         $F_2 = \frac{kq^2}{d^2}$

         $F_2 = \frac{ 9*10^9 * (15 *10^{-6})}{0.025}$

         $F_2 = 3.24*10^{3} \ N$