Question

Two small nonconducting spheres have a total charge of 90.0 C.

(a) When placed 28.0 cm apart, the force each exerts on the other is 12.0 N and is repulsive. What is the charge on each?
(b) What if the force were attractive?

Answer 1

Answer: (a) Smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$.

(b) Smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$.

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

        $Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

                      = $9 \times 10^{-5} C$

Therefore, force between the two spheres will be calculated as follows.

        F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

       12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

       $Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

or,     $Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}$

    $9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}$

    $Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0$

        $Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C$

This means that smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$.

(b)  When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

      $Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

      $Q_{1} - Q_{2} = 9 \times 10^{-5} C$

Now, force between the two spheres is calculated as follows.

    F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

    12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

   $Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

   $Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}$

    $Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}$

        $Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}$

Hence, smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$.

Answer 2

(a). The charges on one sphere is $0.89\mu C$  and $1.18 \mu C$ on other sphere.

(b). If the force were attractive then charge on both sphere will be in opposite sign.

Electrostatic force :

The electrostatic force is given as,

                 $F=k\frac{Q_{1}Q_{2}}{r^{2} }$

Where ,

• $r$ is distance between charges.

• $k$ is coulombs constant,$k=9*10^{9} Nm^{2}/C$

• $Q_{1}$ and $Q_{2}$ are charges on both non conducting spheres  .

Given that, $F=12N,r=28cm=0.28m,Q_{1}+Q_{2}=90$

substitute values in above relation.

             $12=\frac{9*10^{9} *Q_{1}(90-Q_{1})}{(0.28)^{2} } \\\\Q_{1}^{2}-(90*10^{-6})Q_{1} +1.05*10^{-10}=0\\ \\Q_{1}=8.88*10^{-5}=0.89 \mu C\\ \\Q_{2}=1.18*10^{-6} =1.18\mu C$

If the force were attractive then charging on both sphere will be in opposite sign.

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