2025-2026-1 exercise for chapter 21\nstudent id no. 51240661 name abigail odufu\n(1) a negatively - charged…

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