Question

1. when the scholar rubs the balloon on her head for 10 s, the balloon has a charge of -100 e and the hair has a charge of +100 e. if she holds the balloon 1 m from her head, the force between the balloon and her head is about 2 n. which strategy, a or b, would increase the force more? explain your answer.

a. holding the balloon 0.5 m from her head (decreasing the distance between the particles by 2).

b. doubling the charge on her hair to +200 e (increasing the charge between the particles by 2).

Explanation:

Step1: Recall Coulomb's Law

Coulomb's Law is \( F = k\frac{q_1q_2}{r^2} \), where \( F \) is the force, \( k \) is a constant, \( q_1, q_2 \) are charges, and \( r \) is the distance between charges.

Step2: Analyze Strategy A

For Strategy A, original \( r = 1\space m \), new \( r = 0.5\space m \) (so \( r \) is halved, \( r_{new}=\frac{r_{old}}{2} \)). Substituting into Coulomb's Law, \( F \) is inversely proportional to \( r^2 \), so \( F_A = k\frac{q_1q_2}{(r/2)^2}=k\frac{q_1q_2}{r^2/4} = 4\times(k\frac{q_1q_2}{r^2}) \). Original \( F = 2\space N \), so \( F_A = 4\times2 = 8\space N \), change in force \( \Delta F_A = 8 - 2 = 6\space N \).

Step3: Analyze Strategy B

For Strategy B, original \( q_2 = + 100\space e \), new \( q_2 = + 200\space e \) (so charge is doubled, \( q_{2,new}=2q_{2,old} \)). Substituting into Coulomb's Law, \( F \) is directly proportional to the product of charges, so \( F_B = k\frac{q_1(2q_2)}{r^2}=2\times(k\frac{q_1q_2}{r^2}) \). Original \( F = 2\space N \), so \( F_B = 2\times2 = 4\space N \), change in force \( \Delta F_B = 4 - 2 = 2\space N \).

Step4: Compare \( \Delta F_A \) and \( \Delta F_B \)

Since \( 6\space N>2\space N \), Strategy A increases the force more.

Answer:

Strategy A (Holding the balloon 0.5 m from her head) would increase the force more. This is because according to Coulomb's Law \( F = k\frac{q_1q_2}{r^2} \), force is inversely proportional to the square of the distance (so halving the distance quadruples the force) and directly proportional to the product of charges (doubling a charge doubles the force). The increase in force for Strategy A (from 2 N to 8 N, a 6 N increase) is greater than for Strategy B (from 2 N to 4 N, a 2 N increase).