Question

how does the strength of electric force between a pair of charged particles change when the particles are moved twice as close (half the distance)? how will it change if moved five times farther apart? explain and specify numerical values.

Explanation:

Step1: Recall Coulomb's Law

Coulomb's Law for the electric force \( F \) between two charged particles is given by \( F = k\frac{q_1q_2}{r^2} \), where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges of the particles, and \( r \) is the distance between them.

Step2: Analyze when distance is halved (\( r'=\frac{r}{2} \))

Let the initial force be \( F_1 = k\frac{q_1q_2}{r^2} \). When the distance is halved, the new distance \( r_2=\frac{r}{2} \), and the new force \( F_2 = k\frac{q_1q_2}{(r_2)^2}=k\frac{q_1q_2}{(\frac{r}{2})^2}=k\frac{q_1q_2}{\frac{r^2}{4}} = 4k\frac{q_1q_2}{r^2} \). Since \( F_1 = k\frac{q_1q_2}{r^2} \), we have \( F_2 = 4F_1 \). So the force becomes 4 times stronger.

Step3: Analyze when distance is five times farther (\( r' = 5r \))

When the distance is five times farther, the new distance \( r_3 = 5r \), and the new force \( F_3=k\frac{q_1q_2}{(r_3)^2}=k\frac{q_1q_2}{(5r)^2}=k\frac{q_1q_2}{25r^2}=\frac{1}{25}k\frac{q_1q_2}{r^2} \). Since \( F_1 = k\frac{q_1q_2}{r^2} \), we have \( F_3=\frac{1}{25}F_1 \). So the force becomes \( \frac{1}{25} \) of the original force.

Answer:

When the particles are moved twice as close (half the distance), the electric force becomes 4 times stronger. When the particles are moved five times farther apart, the electric force becomes \( \frac{1}{25} \) of its original strength. This is derived from Coulomb's Law \( F = k\frac{q_1q_2}{r^2} \), where force is inversely proportional to the square of the distance between the charged particles.