Question

Three charges q1 = + 6.0 µC, q2 = + 9.0 µC and q3 = + 12.0 µC are located at (0, 0) m, (0.3, 0) m and (0.3, 0.4) m respectively. Find the electric force on q3 due to q1, q3 due to q2, find the net force on q3. Show results in unit vector notation.?

Answer 1

1) Find first the electric force on q₃ due to q₁. Coulomb's law states that the electrostatic force F₁₃ experienced by a charge, q₁ at position r₁, in the vicinity of another charge, q₃ at position r₃, is equal to:

$${{\vec{F}}_{13}}=k{{q}_{1}}{{q}_{3}}\frac{\left( {{{\vec{r}}}_{1}}-{{{\vec{r}}}_{3}} \right)}{{{\left| {{{\vec{r}}}_{1}}-{{{\vec{r}}}_{3}} \right|}^{3}}}\,$$

where constant k is equal to 9·10⁹ N·m²·C^-2. The electrostatic force ${{\vec{F}}_{31}}$  experienced by q₃ due to q₁, according to Newton's third law, is ${{\vec{F}}_{31}}=-{{\vec{F}}_{13}}$ that is

$${{\vec{F}}_{31}}=k{{q}_{1}}{{q}_{3}}\frac{\left( {{{\vec{r}}}_{3}}-{{{\vec{r}}}_{1}} \right)}{{{\left| {{{\vec{r}}}_{3}}-{{{\vec{r}}}_{1}} \right|}^{3}}}\,\,\,\,\,\,\left( 1 \right)$$

We are given

$${{\vec{r}}_{1}}=0\vec{i}+0\vec{j},\,\,\,\,\,{{\vec{r}}_{3}}=0.3\vec{i}+0.4\vec{j}$$

Then

$${{\vec{r}}_{3}}-{{\vec{r}}_{1}}=0.3\vec{i}+0.4\vec{j}$$

$$\left| {{{\vec{r}}}_{3}}-{{{\vec{r}}}_{1}} \right|=\sqrt{{{\left( 0.3 \right)}^{2}}+{{\left( 0.4 \right)}^{2}}}=0.5$$

$${{\left| {{{\vec{r}}}_{3}}-{{{\vec{r}}}_{1}} \right|}^{3}}=0.125$$

Substitute the known values into (1)

$${{{\vec{F}}}_{31}}=9\cdot {{10}^{9}}N\cdot {{m}^{2}}\cdot {{C}^{-2}}\cdot 6.0\cdot {{10}^{-6}}C \\ \cdot 12.0\cdot {{10}^{-6}}C\cdot \frac{0.3\vec{i}+0.4\vec{j}}{0.125\,{{m}^{2}}} \\$$

$${{\vec{F}}_{31}}=5.184\cdot \,\left( 0.3\vec{i}+0.4\vec{j} \right)N=(1.56\vec{i}+2.07\vec{j})N$$

2) Find the electric force on q₃ due to q₂. By analogy with (1), we have

$${{\vec{F}}_{32}}=k{{q}_{2}}{{q}_{3}}\frac{\left( {{{\vec{r}}}_{3}}-{{{\vec{r}}}_{2}} \right)}{{{\left| {{{\vec{r}}}_{3}}-{{{\vec{r}}}_{2}} \right|}^{3}}}\,\,\,\,\,\,\left( 2 \right)$$

We are given

$${{\vec{r}}_{2}}=0.3\vec{i}+0\vec{j},\,\,\,\,\,{{\vec{r}}_{3}}=0.3\vec{i}+0.4\vec{j}$$

Then

$${{\vec{r}}_{3}}-{{\vec{r}}_{2}}=0\vec{i}+0.4\vec{j}$$

$$\left| {{{\vec{r}}}_{3}}-{{{\vec{r}}}_{2}} \right|=0.4$$

$${{\left| {{{\vec{r}}}_{3}}-{{{\vec{r}}}_{2}} \right|}^{3}}=0.064$$

Substitute the known values into (2)

 

$${{{\vec{F}}}_{32}}=9\cdot {{10}^{9}}N\cdot m\cdot {{C}^{-2}}\cdot 9.0\cdot {{10}^{-6}}C \\ \cdot 12.0\cdot {{10}^{-6}}C\cdot \frac{0\vec{i}+0.4\vec{j}}{0.064\,{{m}^{2}}} \\$$

$${{\vec{F}}_{32}}=15.188\cdot \,\left( 0\vec{i}+0.4\vec{j} \right)N=6.08\vec{j}N$$

 

3) Now we find the net force on q₃ applying the superposition principle

$${{\vec{F}}_{3}}={{\vec{F}}_{31}}+{{\vec{F}}_{32}}=(1.56\vec{i}+2.07\vec{j}+6.08\vec{j})N$$

$$=(1.56\vec{i}+8.15\vec{j})N$$