Question

Charges of 3 nC and – 3 nC are placed at the two corners of an equilateral
triangle of side 1 m. Find the resultant electric intensity at the third corner.
[ July - 17]
Ans : 27 NC-1, a = 60°​

Answer 1

In the figure, the electric field due to $\sf{+q,}$

$\longrightarrow\vec{\sf{E_{+q}}}=\sf{\dfrac{kq}{r^2}\left[-\cos60^o\ \hat i+\sin60^o\ \hat j\right]}$

$\longrightarrow\vec{\sf{E_{+q}}}=-\sf{\dfrac{kq}{2r^2}\left[\hat i-\sqrt3\ \hat j\right]}$

And the electric field due to $\sf{-q,}$

$\longrightarrow\vec{\sf{E_{-q}}}=\sf{\dfrac{kq}{r^2}\left[-\cos60^o\ \hat i-\sin60^o\ \hat j\right]}$

$\longrightarrow\vec{\sf{E_{+q}}}=-\sf{\dfrac{kq}{2r^2}\left[\hat i+\sqrt3\ \hat j\right]}$

Hence net electric field,

$\longrightarrow\vec{\sf{E}}=\vec{\sf{E_{+q}}}+\vec{\sf{E_{-q}}}$

$\longrightarrow\vec{\sf{E}}=-\sf{\dfrac{kq}{2r^2}\left[\hat i-\sqrt3\ \hat j\right]-\sf{\dfrac{kq}{2r^2}\left[\hat i+\sqrt3\ \hat j\right]}$

$\longrightarrow\vec{\sf{E}}=-\sf{\dfrac{kq}{r^2}\ \hat i}$

whose magnitude is,

$\longrightarrow\sf{E=\dfrac{kq}{r^2}}$

Here,

• $\sf{k=9\times10^9\ N\,m^2\,C^{-2}}$

• $\sf{q=3\times10^{-9}\ C}$

• $\sf{r=1\ m}$

Then, net electric field,

$\longrightarrow\sf{E=\dfrac{kq}{r^2}}$

$\longrightarrow\sf{E=\dfrac{9\times10^9\times3\times10^{-9}}{1^2}}$

$\longrightarrow\underline{\underline{\sf{E=27\ N\,C^{-1}}}}$