Question

Three point charges of magnitude q are located at the vertices of an equilateral triangle l. What is the charge should be placed at the centroid to keep all the three charges in equilibrium?

Answer 1

Given:

Three point charges of magnitude q are located at the vertices of an equilateral triangle l.

To find:

The charge should be placed at the centroid to keep all the three charges in equilibrium.

Calculation:

Considering a charge at the vertex:

It will be pushed by other vertex charges , but pulled by the centroid charge.

Let charge at Centroid be q_(0).

Pushing force by other vertex charges :

$f1 = \sqrt{ {f}^{2} + {f}^{2} + 2 {f}^{2} \cos( 60 \degree) }$

$= > f1 = \sqrt{ {f}^{2} + {f}^{2} + {f}^{2} }$

$= > f1 = \sqrt{ 3{f}^{2} }$

$= > f1 =f \sqrt{ 3 }$

$= > f1 = \dfrac{k {q}^{2} }{ {l}^{2} } \sqrt{ 3 }$

Pulling force by centroid charge:

$f2 = \dfrac{k (q \times q_{0} )}{ {( \frac{l}{ \sqrt{3} } )}^{2} }$

$= > f2 = \dfrac{3k (q \times q_{0}) }{ {l}^{2} }$

Now, f1 and f2 should equal and opposite:

$\therefore \: f1 = f2$

$= > \dfrac{k {q}^{2} }{ {l}^{2} } \sqrt{3} = \dfrac{3k (q \times q_{0}) }{ {l}^{2} }$

$= > q_{0} = \dfrac{q}{ \sqrt{3} }$

So, final answer is:

$\boxed{ \bold{ \red{ q_{0} = \dfrac{q}{ \sqrt{3} } }}}$