Question

Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two charges of magnitude 2*Q and 2*Q when held at a distance r/2.?

Answer 1

197.5072.

Step-by-step explanation:

According to the Coulomb's law, the magnitude of the electrostatic force of interaction between two charges
`\rm q_1` and
`\rm q_2` which are separated by the distance
`\rm d` is given by

`\rm F = (kq_1q_2)/(d^2).`

where, k is the Coulomb's constant.

For the case, when,

• 
  `\rm q_1 = Q.`
• 
  `\rm q_2 = Q.`
• 
  `\rm d=r.`
• 
  `\rm F=12.3442.`

Then, using Coulomb's law,

`\rm 12.3442 = (kQQ)/(r^2)=(kQ^2)/(r^2)\ \ \ \ .......\ (1).`

For the case, when,

• 
  `\rm q_1 = 2Q.`
• 
  `\rm q_2 = 2Q.`
• 
  `\rm d=\frac r2.`

Then, using Coulomb's law, the new electric force between the charges is given by,

`\rm F' = (k(2Q)(2Q))/(\left (\frac r2\right )^2)\\=(k\ 4Q^2)/((r^2)/(4))\\=4* 4 * (kQ^2)/(r^2)\\=16\ (kQ^2)/(r^2)\\=16* 12.3442\ \ \ \ \ \ \ \ (Using\ (1))\\=197.5072.`