Question

Suppose two point charges $100\mu C$and $5\mu C$are placed at point A and B respectively with AB =$40cm$. Find the work done by external force in displacing the charge $5\mu C$ from B to C , Where BC =$30cm$, angel $ABC=\dfrac{\pi }{2}$ and $\dfrac{1}{4\pi \varepsilon \circ }=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}$.
A. $9J$
B. $\dfrac{81}{20}J$
C. $\dfrac{9}{25}J$
D. $\dfrac{9}{4}J$

Answer 1

Work is a product of measure of energy or charge transfer by an external force in the direction of its placement. We used the formula of work done to solve this question.

Complete step-by-step solution:
As , we know when the charge is moved from one point to another, work is $W=q(VA-VB)$ , here W= denotes work done.
q= charge
$VA$= electric potential at point A
$VB$= electric potential at point
Here , work done by external force in displacing the charge $5\mu C$ from B to C is $W=5\times {{10}^{-5}}(VC-VB)$
Let's draw a diagram of this problem to solve this problem easily-

Now we have to calculate the values of $VA$and $VB$ -

& VB\Rightarrow 9\times {{10}^{9}}\times \dfrac{100\times {{10}^{-6}}}{0.4} \\\ & VB\Rightarrow \dfrac{9}{4}\times {{10}^{6}} \\\ \end{aligned}$$ And $$\begin{aligned} & VC\Rightarrow 9\times {{10}^{9}}\times \dfrac{100\times {{10}^{-6}}}{0.5} \\\ & VC\Rightarrow \dfrac{9}{5}\times {{10}^{6}} \\\ \end{aligned}$$ So , work done is $$\begin{aligned} & \Rightarrow W=5\times {{10}^{-6}}\times \left( \dfrac{9}{5}\times {{10}^{6}}-\dfrac{9}{4}\times {{10}^{6}} \right) \\\ & \Rightarrow W=5\times {{10}^{-6}}\times \left( \dfrac{9}{5}-\dfrac{9}{4} \right){{10}^{6}} \\\ & \Rightarrow W=5\times {{10}^{-6}}\times \left( \dfrac{9}{20} \right){{10}^{6}} \\\ & \Rightarrow W=5\times \dfrac{9}{20} \\\ & \Rightarrow W=\dfrac{9}{4} \\\ \end{aligned}$$ Therefore, correct option is D. $$\dfrac{9}{4}J$$ **Note:** Note that the work becomes $$(-)ve$$ , when the force and displacement are in opposite directions to each other. The work done by force is null when the direction of force and displacement are perpendicular to each other.