Question

how far apart are the two charges one of magnitude 3uc and the other of magnitude 5uc if the force between them is 20N

Answer 1

Answer:

Let the force Initially the force between them is will be (k*5*10)*10^-12/r^2=F1

when -4uC is added to both then the force between them will be 5–1=1; 10–4=6; the force between them will be (k*1*6)*10^-12/r^2=F2

now divide both equations ie, F1/F2= 5*10/6= {F1=25/3}

Answer 2

The two charges q1 and q2 have a  distance of $8.21 * 10^{-2} m$ or 8.21 cm between each other.

Coulomb's Law for charges :

• According to Coulomb, two charges placed at some distance between each other, exert a force between them.
• The force is of attraction if the charges are of different types i.e. positive and negative or vice versa.
• The force is of repulsion if the charges are of the same sign i.e. both positive or both negative.

The formulation according to Coulomb for the force between two charges 'q1' and 'q2' placed at distance 'r' apart is :

              $F = k q_{1}q_{2} / r^{2}$

              $k = 8.99 * 10^{9} Nm^{2}/C^{2}$

Thus, we have the following values ( in standard units ) :

Charge of the first particle : $q_{1} = 3uc = 3 * 10^{-6} C$

Charge of the second particle : $q_{2} = 5uc = 5 * 10^{-6} C$

The force between the two particles : $F = 20N$

So, applying Coulomb's formula :

                 $r^{2} = k q_{1}q_{2} / F\\r^{2} = k * (3*10^{-6} )*(5*10^{-6} ) / 20\\r^{2} = k * (15*10^{-12} )/ 20\\r^{2} = k * 0.75 10^{-12} \\r^{2} = 8.99 * 10^{9} * 0.75* 10^{-12} \\r^{2} = 6.7425 * 10^{9-12}\\r^{2} = 6.7425 * 10^{-3}\\r^{2} = 67.425 * 10^{-4}$

Taking square root on both sides to get the distance 'r' :

                 $r^{2} = 67.425 * 10^{-4}\\\sqrt{ r^{2}} = \sqrt{ 67.425 * 10^{-4}}\\r =10^{-2}} *\sqrt{ 67.425 }\\ r = 8.21 * 10^{-2}$

Hence, the two charges are at a distance of $8.21 * 10^{-2} m$ from each other.

                     

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