ElectrostaticsMedium

Question

Two point charges $q_{1} = 3\mu C$ and $q_{2} = - 4\mu C$ are placed at points $(2\widehat{i} + 3\widehat{j} + 3\widehat{k})$ and $(\widehat{i} + \widehat{j} + \widehat{k})$ respectively. Force on charge $q_{2}$ is $\_\_\_\_$ N.

(Take $\frac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9}$ SI Units)

Options

A.$(12\widehat{i} + 24\widehat{j} + 24\widehat{k}) \times 10^{- 3}$
B.$(4\widehat{i} + 8\widehat{j} + 8\widehat{k}) \times 10^{- 3}$
C.$(3\widehat{i} + 6\widehat{j} + 6\widehat{k}) \times 10^{- 3}$
D.$( - 4\widehat{i} - 8\widehat{j} - 8\widehat{k}) \times 10^{- 3}$

Solution

Sol. [Key Concept: Coulomb force $F = \frac{{kq}_{1}q_{2}}{r^{3}}\ r$ with the separation vector $r = r_{2} - r_{1}$; evaluating gives $(4i + 8j + 8k) \times 10^{- 3}$ N.]

$$\overrightarrow{r} = {\overrightarrow{r}}_{2} - {\overrightarrow{r}}_{1} \Rightarrow - \widehat{i} - 2\widehat{j} - 2\widehat{k} $$Force $\Rightarrow \frac{{kq}_{1}q_{2}}{r^{3}}\overrightarrow{r}$

$${\Rightarrow \frac{9 \times 10^{9} \times (3)( + 4) \times 10^{- 12}}{27}(\widehat{i} + 2\widehat{j} + 2\widehat{k}) }{\Rightarrow 4 \times 10^{- 3}\lbrack\widehat{i} + 2\widehat{j} + 2\widehat{k}\rbrack }$$

Create a free account to view solution

View Solution Free
Topic: Electrostatics·Practice all Electrostatics questions

More Electrostatics Questions