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

Answer:

197.5072.

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

$\rm F = \dfrac{kq_1q_2}{d^2}.$

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,

$\rm 12.3442 = \dfrac{kQQ}{r^2}=\dfrac{kQ^2}{r^2}\ \ \ \ .......\ (1).$

For the case, when,

• $\rm q_1 = 2Q.$
• $\rm q_2 = 2Q.$
• $\rm d=\dfrac r2.$

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

$\rm F' = \dfrac{k(2Q)(2Q)}{\left (\dfrac r2\right )^2}\\=\dfrac{k\ 4Q^2}{\dfrac{r^2}{4}}\\=4\times 4 \times \dfrac{kQ^2}{r^2}\\=16\ \dfrac{kQ^2}{r^2}\\=16\times 12.3442\ \ \ \ \ \ \ \ (Using\ (1))\\=197.5072.$