Question

1. three identical point charges of 2.0 μc are placed on the x - axis. the first charge is at the origin, the second to the right at x = 50 cm, and the third is at the 100 cm mark. what are the magnitude and direction of the electrostatic force which acts on the charge at the origin?

a. 0.18 n, right
b. 0.18 n, left
c. 0.36 n, right
d. 0.36 n, left

1. two charged particles (-40.0×10^{-6}c and +108×10^{-6}c) are a distance r apart. the electrostatic force they exert on each other is 4.00 n. calculate the value of r.

a. 2.10 m
b. 3.67 m
c. 2.49 m
d. 3.12 m

Explanation:

Question 21

Step1: Calculate the force due to the charge at \(x = 50\ cm\)

Coulomb's law is \(F=\frac{kq_1q_2}{r^{2}}\). Here, \(k = 9\times10^{9}\ N\cdot m^{2}/C^{2}\), \(q_1=q_2 = 2\times10^{- 6}\ C\), and \(r = 0.5\ m\).
\(F_1=\frac{9\times10^{9}\times(2\times10^{-6})\times(2\times10^{-6})}{(0.5)^{2}}\)

$$
LATEXBLOCK0
$$

Since like - charges repel, the force \(F_1\) on the charge at the origin due to the charge at \(x = 50\ cm\) is to the left.

Step2: Calculate the force due to the charge at \(x = 100\ cm\)

Here, \(r = 1\ m\)
\(F_2=\frac{9\times10^{9}\times(2\times10^{-6})\times(2\times10^{-6})}{(1)^{2}}\)

$$
LATEXBLOCK1
$$

Since like - charges repel, the force \(F_2\) on the charge at the origin due to the charge at \(x = 100\ cm\) is to the left.

Step3: Calculate the net force

\(F_{net}=F_1 + F_2\)
\(F_{net}=0.144+0.036=0.18\ N\)
The direction is to the left (because both forces \(F_1\) and \(F_2\) are to the left)

Step1: Use Coulomb's law \(F=\frac{k|q_1q_2|}{r^{2}}\)

We know that \(F = 4\ N\), \(k=9\times10^{9}\ N\cdot m^{2}/C^{2}\), \(q_1=-40\times10^{-6}\ C\), \(q_2 = 108\times10^{-6}\ C\)
First, calculate \(|q_1q_2|=(40\times10^{-6})\times(108\times10^{-6}) = 4320\times10^{-12}\ C^{2}\)
From \(F=\frac{k|q_1q_2|}{r^{2}}\), we can solve for \(r\):
\(r^{2}=\frac{k|q_1q_2|}{F}\)
Substitute the values:

$$
LATEXBLOCK0
$$

\(r=\sqrt{9.72}\approx3.12\ m\)

Answer:

B. \(0.18\ N\), left

Question 22