Two identical metal spheres are placed 0.2 m apart. A charge (q1) of 9 μC is placed on one sphere while a charge (q2) of -3 μC is placed upon the other. (a) What is the force on ea... Two identical metal spheres are placed 0.2 m apart. A charge (q1) of 9 μC is placed on one sphere while a charge (q2) of -3 μC is placed upon the other. (a) What is the force on each of the spheres? (b) If the two spheres are brought together and touched and then returned to their original positions, what will be the force on each sphere? Read more

Steps to Solve

1. Identify the charges and distance
   Given charges are ( q_1 = 9 , \mu C = 9 \times 10^{-6} , C ) and ( q_2 = -3 , \mu C = -3 \times 10^{-6} , C ).
   The distance ( r ) between the spheres is ( 0.2 , m ).

2. Apply Coulomb's Law
   Coulomb's Law states that the electric force ( F ) between two charges is given by
   $$ F = k \frac{|q_1 \cdot q_2|}{r^2} $$
   where ( k ) is Coulomb's constant ( (8.99 \times 10^9 , N m^2/C^2) ).

3. Calculate the magnitude of the force
   Substituting the values into the equation, we get:
   $$ F = 8.99 \times 10^9 , \frac{(9 \times 10^{-6}) \cdot (-3 \times 10^{-6})}{(0.2)^2} $$
   Calculating this gives:
   $$ F = 8.99 \times 10^9 \times \frac{27 \times 10^{-12}}{0.04} $$

4. Compute the force value
   Continuing the calculation:
   $$ F = 8.99 \times 10^9 \times 675 \times 10^{-12} $$
   $$ F \approx 6.06125 , N $$
   Since charge ( q_1 ) is positive and ( q_2 ) is negative, the force is attractive.

5. Determine the effect of touching the spheres
   When the spheres are touched, the total charge ( q_{total} = 9 , \mu C - 3 , \mu C = 6 , \mu C ) is evenly distributed. Each sphere will then have:
   $$ q' = \frac{q_{total}}{2} = \frac{6 , \mu C}{2} = 3 , \mu C $$

6. Calculate the new force after separating
   Using the same formula with the new charges:
   $$ F' = k \frac{|3 \times 10^{-6} \cdot 3 \times 10^{-6}|}{(0.2)^2} $$
   Calculating, we get:
   $$ F' = 8.99 \times 10^9 \frac{9 \times 10^{-12}}{0.04} $$
   Continuing:
   $$ F' = 8.99 \times 10^9 \times 225 \times 10^{-12} $$

7. Final computation of the new force value
   Calculating gives:
   $$ F' \approx 2.022375 , N $$