Question 271: Electrostats & Capacitance

Question

Two point charges of 1 nC and 2 nC are placed at the two corners of equilateral triangle of side 3 cm . The work done in bringing a charge of 3 nC from infinity to the third corner of the triangle is $\_\_\_\_$ $\mu \mathrm{J}$.
$$ \frac{1}{4 \pi \epsilon_0}=9 	imes 10^9 \mathrm{~N} \cdot \mathrm{~m}^2 / \mathrm{C}^2 $$

Accepted Answer

When 3 nC is at infinity then electric potential due to charges 1 nC and 2 nC is zero. So the initial potential is $\mathrm{V}_{\mathrm{i}}$
The total potential at the third corner due to $\mathrm{q}_1(1 \mathrm{nC})$ and $\mathrm{q}_2(2 \mathrm{nC})$ is the sum of their individual potentials:
$$ \mathrm{V}_{\mathrm{f}}=\frac{\mathrm{kq}_1}{\mathrm{r}}+\frac{\mathrm{kq}_2}{\mathrm{r}}=\frac{\mathrm{k}\left(\mathrm{q}_1+\mathrm{q}_2ight)}{\mathrm{r}} $$
The work done is given by $\mathrm{W}=\mathrm{q}_3 	imes \Delta \mathrm{V}$, where $\Delta \mathrm{V}=$ potential difference between initial and final position.
$$ \mathrm{W}=\mathrm{q}_3 	imes\left[\frac{\mathrm{k}\left(\mathrm{q}_1+\mathrm{q}_2ight)}{\mathrm{r}}-0ight] $$
$$\Rightarrow $$ $$ \mathrm{W}=\frac{\mathrm{k} 	imes \mathrm{q}_3 	imes\left(\mathrm{q}_1+\mathrm{q}_2ight)}{\mathrm{r}} $$
Substituting the values,
$$ \mathrm{W}=\frac{\left(9 	imes 10^9ight) 	imes\left(3 	imes 10^{-9}ight) 	imes\left(1 	imes 10^{-9}+2 	imes 10^{-9}ight)}{0.03} $$
$$\Rightarrow $$ $$ \mathrm{W}=\frac{(27) 	imes\left(3 	imes 10^{-9}ight)}{0.03}=27 	imes 10^{-7} \mathrm{~J} $$
As $1 \mathrm{~J}=10^6 \mu \mathrm{~J}$
$$ \mathrm{W}=27 	imes 10^{-7} 	imes 10^6 \mu \mathrm{~J} $$
$$\Rightarrow $$ $$ W=27 	imes 10^{-1} \mu J=2.7 \mu J $$
So, the work done in bring a charge of 3 nC from infinity to the third corner of the triangle is $2.7 \mu \mathrm{~J}$.
Hence, the correct option is (C).

Answer

3.3

Answer

27

Answer

2.7

Answer

5.4

Electrostats & Capacitance

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