Question 135: Electrostatics & Capacitance

Question

Two point charges $\mathrm{q}_1=3 \mu C$ and $\mathrm{q}_2=-4 \mu C$ are placed at points $(2 \hat{i}+3 \hat{j}+3 \hat{k})$ and $(\hat{i}+\hat{j}+\hat{k})$ respectively. Force on charge $\mathrm{q}_2$ is $\_\_\_\_$ N. (Take $\frac{1}{4 \pi \epsilon_0}=9 	imes 10^9$ SI Units)

Accepted Answer

Coulomb's Law states that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
In vector form, the force $\vec{F}_{21}$ exerted on charge $q_2$ by charge $q_1$ is:
$$ \overrightarrow{\mathrm{F}}_{21}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{|\overrightarrow{\mathrm{r}}|^2} \hat{\mathrm{r}} $$
Where,
$\vec{r}=\vec{r}_2-\vec{r}_1$ is the position vector from $q_1$ to $q_2, \hat{r}=\frac{\vec{r}}{|\vec{r}|}$ is the unit vector and $\frac{1}{4 \pi \epsilon_0}=9 	imes 10^9$
Substituting the unit vector, we get:
$$ \overrightarrow{\mathrm{F}}_{21}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{|\overrightarrow{\mathrm{r}}|^3} \overrightarrow{\mathrm{r}} $$
Let $\vec{r}_1$ be the position of $q_1$ and $\vec{r}_2$ be the position of $q_2$ :
$$ \overrightarrow{\mathrm{r}}_1=2 \hat{\imath}+3 \hat{\jmath}+3 \hat{\mathrm{k}} $$
$$ \overrightarrow{\mathrm{r}}_2=1 \hat{\imath}+1 \hat{\jmath}+1 \hat{\mathrm{k}} $$
The vector from $\mathrm{q}_1$ to $\mathrm{q}_2$ is :
$$ \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{r}}_2-\overrightarrow{\mathrm{r}}_1=(1-2) \hat{\mathrm{i}}+(1-3) \hat{\mathrm{j}}+(1-3) \hat{\mathrm{k}} $$
$$\Rightarrow $$ $$ \overrightarrow{\mathrm{r}}=-\hat{\imath}-2 \hat{\jmath}-2 \hat{\mathrm{k}} $$
So, the magnitude of position vector is,
$$ |\overrightarrow{\mathrm{r}}|=\sqrt{(-1)^2+(-2)^2+(-2)^2} $$
$$\Rightarrow $$ $$ |\overrightarrow{\mathrm{r}}|=\sqrt{1+4+4}=\sqrt{9}=3 \mathrm{~m} $$
Substituting the values into the vector formula :
$\vec{F}=\left(9 	imes 10^9ight) \frac{\left(3 	imes 10^{-6}ight)\left(-4 	imes 10^{-6}ight)}{3^3}(-\hat{\imath}-2 \hat{\jmath}-2 \hat{k})$
$$\Rightarrow $$ $\overrightarrow{\mathrm{F}}=\frac{9 	imes 3 	imes(-4) 	imes 10^{-3}}{27}(-\hat{\imath}-2 \hat{\jmath}-2 \hat{k})$
$$\Rightarrow $$ $\vec{F}=\frac{-108 	imes 10^{-3}}{27}(-\hat{\imath}-2 \hat{\jmath}-2 \hat{k})$
$$\Rightarrow $$ $\overrightarrow{\mathrm{F}}=-4 	imes 10^{-3}(-\hat{\imath}-2 \hat{\jmath}-2 \hat{\mathrm{k}})$
$$\Rightarrow $$ $$ \overrightarrow{\mathrm{F}}=(4 \hat{\imath}+8 \hat{\jmath}+8 \hat{\mathrm{k}}) 	imes 10^{-3} \mathrm{~N} $$
Therefore, the force on charge $\mathrm{q}_2$ is $(4 \hat{\imath}+8 \hat{\jmath}+8 \hat{\mathrm{k}}) 	imes 10^{-3} \mathrm{~N}$.
Hence, the correct option is (B).

Answer

$(12 \hat{i}+24 \hat{j}+24 \hat{k}) 	imes 10^{-3}$

Answer

$(4 \hat{i}+8 \hat{j}+8 \hat{k}) 	imes 10^{-3}$

Answer

$(3 \hat{i}+6 \hat{j}+6 \hat{k}) 	imes 10^{-3}$

Answer

$(-4 \hat{i}-8 \hat{j}-8 \hat{k}) 	imes 10^{-3}$

Electrostatics & Capacitance

Two point charges \mathrm{q}_1=3 \mu C and \mathrm{q}_2=-4 \mu C are placed at points (2 \hat{i}+3 \hat{j}+3 \hat{k}) and (\hat{i}+\hat{j}+\hat{k}) respectively. Force on charge \mathrm{q}_2 is \_\_\_\_ N. (Take \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 SI Units)

Choose the correct answer: