Question

honors physics
unit 2: forces at a distance
a negative charge of -8.1x10^6 c exerts an attractive force of 75 n on a second charge that is 0.040 m away. how much charge does the second charge have?
k=
q1=
n=
r=
a negative charge of -5.2x10^6 c exerts an attractive force of 25 n on a second charge that is 0.15 m away. how much charge does the second charge have?
k=
q1=
n=
r=
a negative charge of -1.6x10^6 c exerts an attractive force of 39 n on a second charge that is 0.10 m away. how much charge does the second charge have?
k=
q1=
n=
r=
a negative charge of -2.1x10^6 c exerts an attractive force of 105 n on a second charge that is 0.040 m away. how much charge does the second charge have?
k=
q1=
n=
r=
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Explanation:

Step1: Write Coulomb's Law formula

Coulomb's Law is \(F = k\frac{\vert q_1q_2\vert}{r^{2}}\). We need to solve for \(q_2\). Rearranging the formula gives \(q_2=\frac{F r^{2}}{k\vert q_1\vert}\). The value of \(k = 9\times10^{9}\ N\cdot m^{2}/C^{2}\).

Step2: Substitute values for the first problem

For the first problem: \(q_1=- 8.1\times10^{-6}\ C\), \(F = 75\ N\), \(r = 0.040\ m\).
\(\vert q_1\vert=8.1\times 10^{-6}\ C\), \(r^{2}=(0.040)^{2}=0.0016\ m^{2}\)
\(q_2=\frac{75\times0.0016}{9\times10^{9}\times8.1\times10^{-6}}\)

$$
LATEXBLOCK0
$$

Step3: Substitute values for the second problem

For the second problem: \(q_1=-5.2\times10^{-6}\ C\), \(F = 25\ N\), \(r = 0.15\ m\)
\(\vert q_1\vert = 5.2\times10^{-6}\ C\), \(r^{2}=(0.15)^{2}=0.0225\ m^{2}\)
\(q_2=\frac{25\times0.0225}{9\times10^{9}\times5.2\times10^{-6}}\)

$$
LATEXBLOCK1
$$

Step4: Substitute values for the third problem

For the third problem: \(q_1=-1.6\times10^{-6}\ C\), \(F = 39\ N\), \(r = 0.10\ m\)
\(\vert q_1\vert=1.6\times10^{-6}\ C\), \(r^{2}=(0.10)^{2}=0.01\ m^{2}\)
\(q_2=\frac{39\times0.01}{9\times10^{9}\times1.6\times10^{-6}}\)

$$
LATEXBLOCK2
$$

Step5: Substitute values for the fourth problem

For the fourth problem: \(q_1=-2.1\times10^{-6}\ C\), \(F = 105\ N\), \(r = 0.040\ m\)
\(\vert q_1\vert=2.1\times10^{-6}\ C\), \(r^{2}=(0.040)^{2}=0.0016\ m^{2}\)
\(q_2=\frac{105\times0.0016}{9\times10^{9}\times2.1\times10^{-6}}\)

$$
LATEXBLOCK3
$$

Answer:

For the first problem: \(q_2 = 1.65\times10^{-6}\ C\)
For the second problem: \(q_2=1.2\times10^{-5}\ C\)
For the third problem: \(q_2 = 2.71\times10^{-5}\ C\)
For the fourth problem: \(q_2=8.89\times10^{-6}\ C\)