Question

hey mates...


Answer my previous question too.....

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Answer 1

Answer:

Given:

3 charges of equal magnitude are placed at the 3 vertices of an equilateral triangle of side l

To find:

Force experienced by a. charge placed at the centroid.

Concept:

First of all you need to calculate the distance of the charges from the centroid.

Altitude in an equilateral triangle is as follows :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{d = \dfrac{ \sqrt{3} }{2} l}$

Centroid divides this line in the

ratio of 2 : 1 .

So distance of each charge (at vertex) from the centroid is :

$r = \dfrac{2}{3} \times \dfrac{ \sqrt{3} }{2} l = \dfrac{l}{ \sqrt{3} }$

Calculation:

Force experienced by the centre charge due to each vertex charge is :

$f = \dfrac{1}{4\pi \epsilon} ( \dfrac{qQ}{ {r}^{2} } )$

$= > f = \dfrac{1}{4\pi \epsilon} \{\dfrac{qQ}{ { (\frac{l}{ \sqrt{ 3} } )}^{2} } \}$

$= > f = \dfrac{3}{4\pi \epsilon} \{ \dfrac{qQ}{ {l}^{2} } \}$

Now these forces are vector quantities which are located at 120° to each other.

When equal vectors are located at 120° to each other , the net resultant will be zero.

So final answer :

$\boxed{ \large{ \sf{ \red{f_{net} = 0 \: Newton}}}}$

Answer 2

Answer:

hello

plz mark my answers as brainlist