Question

An electrostatic force of attraction between two point charges A and B is $1000\,{\text{N}}$. If the charge on A is increased by 25% and that on B is reduced by 25% and the initial distance between them is decreased by 25%, the new force of attraction between them is ________N.
A. 1666.67
B. 3256.33
C. 1253.45
D. 3333.3

Answer 1

Use the expression for Coulomb’s law of electric charges. This expression gives the relation between the electrostatic force of attraction or repulsion between two charges, values of two charges and distance between these charges. Determine the values of new charges and distance according to given information and solve it for the new force.

Formula used:
The expression for Coulomb’s law of electric charges is
$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ …… (1)
Here, $F$ is the electrostatic force of attraction or repulsion between charges, $k$ is the constant, ${q_1}$ is the first charge, ${q_2}$ is the second charge and $r$ is the distance between the two charges.

Complete step by step solution:
We have given that there is an electrostatic force of attraction between the charges A and B.
The equation (1) for the electrostatic force of attraction between charges A and B becomes
$\Rightarrow F = k\dfrac{{{q_A}{q_B}}}{{{r^2}}}$ …… (2)
Here, ${q_A}$ is the charge on A and ${q_B}$ is the charge on B.
The charge on A is increased by 25%. Hence, the new charge ${q_A}'$ becomes
${q_A}' = \left( {1 + \dfrac{1}{4}} \right){q_A}$
$\Rightarrow {q_A}' = \dfrac{5}{4}{q_A}$
The charge on A is reduced by 25%. Hence, the new charge ${q_B}'$ becomes
$\Rightarrow{q_B}' = \left( {1 - \dfrac{1}{4}} \right){q_B}$
$\Rightarrow {q_B}' = \dfrac{3}{4}{q_B}$
The distance between the two charges A and B is decreased by 25%. Hence, the new distance $r'$ between charges becomes
$\Rightarrow r' = \left( {1 - \dfrac{1}{4}} \right)r$
$\Rightarrow r' = \dfrac{3}{4}r$
Now the equation (1) for the new electrostatic force $F'$ of attraction between the new charges is
$\Rightarrow F' = k\dfrac{{{q_A}'{q_B}'}}{{r{'^2}}}$
Substitute $\dfrac{5}{4}{q_A}$ for ${q_A}'$, $\dfrac{3}{4}{q_B}$ for ${q_B}'$ and $\dfrac{3}{4}r$ for $r'$ in the above equation.
$\Rightarrow F' = k\dfrac{{\left( {\dfrac{5}{4}{q_A}} \right)\left( {\dfrac{3}{4}{q_B}} \right)}}{{{{\left( {\dfrac{3}{4}r} \right)}^2}}}$
$\Rightarrow F' = k\dfrac{{15{q_A}{q_B}}}{{9{r^2}}}$ …… (3)
Divide equation (3) by equation (2).
$\Rightarrow \dfrac{{F'}}{F} = \dfrac{{k\dfrac{{15{q_A}{q_B}}}{{9{r^2}}}}}{{k\dfrac{{{q_A}{q_B}}}{{{r^2}}}}}$
$\Rightarrow \dfrac{{F'}}{F} = \dfrac{{15}}{9}$
Rearrange the above equation for $F'$.
$\Rightarrow F' = \dfrac{{15}}{9}F$
Substitute $1000\,{\text{N}}$ for $F$ in the above equation.
$\Rightarrow F' = \dfrac{{15}}{9}\left( {1000\,{\text{N}}} \right)$
$\therefore F' = 1666.67\,{\text{N}}$
Therefore, the new force of attraction between the charges is $1666.67\,{\text{N}}$.

Hence, the correct option is A.

Note: The students should be careful while determining the values of the new charges and distance between them. If these values are not determined correctly, the final answer will also be incorrect. Also, the value of k remains the same for both electrostatic forces of attraction as it is a constant.