Solution:

Ans. (i) The magnitude

$$ \begin{align*} & \left|\overrightarrow{E_{A B}}\right|=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{a^{2}}=E \ & \left|\overrightarrow{E_{A C}}\right|=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 q}{a^{2}}=2 E \end{align*} $$

$$ \begin{aligned} & E_{\text {net }}=\sqrt{4 E^{2}+E^{2}-2 E^{2}} \ & E_{\text {net }}=E \sqrt{3}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q \sqrt{3}}{a^{2}} \end{aligned} $$

(ii) The direction of resultant electric field at vertex $A$

$$ \begin{align*} \tan \alpha & =\frac{E_{A B} \sin 120^{\circ}}{E_{A C}+E_{A B} \cos 120^{\circ}} \ \tan \alpha & =\frac{E \times \frac{\sqrt{3}}{2}}{2 E+E \times\left(\frac{-1}{2}\right)}=\frac{1}{\sqrt{3}} \ \alpha & \left.=30^{\circ} \text { (with side } A C\right) \end{align*} $$

[CBSE Marking Scheme, 2014]

[AT Q. 7. A charge is distributed uniformly over a ring of radius ’ $a$ ‘. Obtain an expression for the electric intensity $E$ at a point on the axis of the ring. Hence, show that for points at large distances from the ring, it behaves like a point charge.

A [Delhi I, II, III, 2016]

$O P=\sqrt{a^{2}+x^{2}}$, by Pythagoras theorem

where, $a$ is radius of ring

$x$ is the distance from centre to point ’ $P$ '

$$ \begin{aligned} d E & =\frac{k d Q}{a^{2}+x^{2}} \ d E_{x} & =\frac{k d Q}{\left(a^{2}+x^{2}\right)} \cos \theta \end{aligned} $$

and $d E_{y}$ will be zero because all components will cancel out each other.

$$ \begin{aligned} \cos \theta & =\frac{x}{\left(a^{2}+x^{2}\right)^{1 / 2}} \ d E_{x} & =\frac{k d Q}{a^{2}+x^{2}} \cdot \frac{x}{\left(a^{2}+x^{2}\right)^{1 / 2}} \ E_{x} & =\int \frac{k x d Q}{\left(a^{2}+x^{2}\right)^{3 / 2}} \ E_{x} & =\frac{k x}{\left(a^{2}+x^{2}\right)^{3 / 2}} \int d Q \ E & =\frac{k Q x}{\left(a^{2}+x^{2}\right)^{3 / 2}} \end{aligned} $$

Now, if $x»a$

$E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{x^{2}}$, which is equal to field produced by a point change.

$\therefore$ For large distance, ring behaves as a point charge. 1

[CBSE Marking Scheme, 2016]