Three equal charges +q are placed at the three vertices of an equ… - Vidyadip

MCQ

Three equal charges $+q$ are placed at the three vertices of an equilateral triangle centred at the origin. They are held in equilibrium by a restoring force of magnitude $F(r)=k r$ directed towards the origin, where $k$ is a constant. What is the distance of the three charges from the origin?

A
$\left[\frac{1}{6 \pi \varepsilon_0} \frac{q^2}{k}\right]^{1 / 2}$
✓
$\left[\frac{\sqrt{3}}{12 \pi \varepsilon_0} \frac{q^2}{k}\right]^{1 / 3}$
C
$\left[\frac{1}{6 \pi \varepsilon_{0}} \frac{q^2}{k}\right]^{2 / 3}$
D
$\left[\frac{\sqrt{3}}{4 \pi \varepsilon_0} \frac{q^2}{k}\right]^{2 / 3}$

✓

Answer

Correct option: B.

$\left[\frac{\sqrt{3}}{12 \pi \varepsilon_0} \frac{q^2}{k}\right]^{1 / 3}$

b
(b)

In given charge configuration,

Net force on any of charge is

$F_{\text {net }}=\sqrt{F_A^2+F_B^2+2 F_A F_B \cos 60^{0}}$

$=\frac{\sqrt{3} \cdot q^2}{4 \pi \varepsilon_{0} a^2}$

where, $a=$ side length of equilateral triangle.

So, radius $r$ is $r=\frac{2}{3}\left(\frac{\sqrt{3}}{2} a\right)$

$\Rightarrow \quad a=\sqrt{3} r$

$\text { Hence, } F_{\text {net }}=\frac{\sqrt{3} \cdot q^2}{\left(4 \pi \varepsilon_0\right) 3 r^2}$

or $F_{\text {net }}=\frac{q^2}{\left(4 \pi \varepsilon_0\right) \sqrt{3} r^2}$

Now, given that this force is balanced by a force $F(r)=k r$

$\therefore k r =\frac{q^2}{\left(4 \pi \varepsilon_0\right) \sqrt{3} r^2}$

$\Rightarrow r^3=\frac{\sqrt{3} q^2}{12 \pi \varepsilon_0 k}$

So, $\quad r=\left(\frac{\sqrt{3} q^2}{12 \pi \varepsilon_0 k}\right)^{\frac{1}{3}}$

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