Question

2025-2026-1 exercise for chapter 21
student id no. 51240661 name abigail odufu
(1) a negatively - charged rod is brought near one end of an uncharged metal bar. the end at the metal bar farthest from the charged rod will be charged (
(a) positive. (b) negative. (c) neutral (d) none of the given answers
(2) an originally neutral electroscope is grounded briefly while a positively charged glass rod is held near it. after the glass rod is removed, the electroscope (
(a) remains neutral.
(b) is negatively charged.
(c) is positively charged.
(d) could be either positively or negatively charged, depending on how long the contact with ground lasted.
(3) the force of attraction between a - 40.0 μc and + 108 μc charge is 4.00 n. what is the separation between these two charges? (
(a) 1.13 m (b) 2.10 m (c) 3.12 m (d) 3.67 m
(4) two point charges, initially 2.0 cm apart, experience a 1.0 - n force. if they are moved to a new separation of 8.0 cm, what is the electric force between them? (
(a) 4.0 n (b) 16 n (c) 1/4 n (d) 1/16 n

Explanation:

Step1: Understand electrostatic induction

When a negatively - charged rod is brought near an uncharged metal bar, electrons in the metal bar are repelled to the far - end. So the end of the metal bar farthest from the charged rod will be negatively charged.

Step2: Analyze grounding and charging of electroscope

When a positively charged glass rod is held near a grounded electroscope, electrons flow from the ground to the electroscope. After the glass rod is removed, the electroscope is negatively charged.

Step3: Use Coulomb's law $F = k\frac{q_1q_2}{r^{2}}$

Given $F = 4.00N$, $q_1=- 40.0\times10^{-6}C$, $q_2 = 108\times10^{-6}C$, $k = 9\times10^{9}N\cdot m^{2}/C^{2}$. Rearranging for $r$, we have $r=\sqrt{k\frac{\vert q_1q_2\vert}{F}}=\sqrt{\frac{9\times10^{9}\times40.0\times10^{-6}\times108\times10^{-6}}{4.00}}=\sqrt{\frac{9\times40.0\times108\times10^{-3}}{4.00}}=\sqrt{9\times108\times10^{-2}}=\sqrt{9.72}\approx3.12m$

Step4: Use Coulomb's law again

Coulomb's law is $F = k\frac{q_1q_2}{r^{2}}$. The initial force $F_1 = k\frac{q_1q_2}{r_1^{2}} = 1.0N$ with $r_1 = 2.0cm$. The new distance $r_2 = 8.0cm$. Since $F_2=k\frac{q_1q_2}{r_2^{2}}$, and $\frac{F_2}{F_1}=\frac{r_1^{2}}{r_2^{2}}$. Substituting $r_1 = 2.0cm$ and $r_2 = 8.0cm$, we get $\frac{F_2}{F_1}=\frac{2.0^{2}}{8.0^{2}}=\frac{4}{64}=\frac{1}{16}$, so $F_2=\frac{1}{16}N$

Answer:

(1) B. negative
(2) B. is negatively charged
(3) C. 3.12 m
(4) D. 1/16 N