Question

Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. below *
2 points
Captionless Image
3N
Zero
15Qq/l
none of the above

Answer 1

Given:

q1, q2 and q3 three charges placed at the vertices of an equilateral triangle of side l. A charge Q is placed at the centroid of the triangle.

To find:

Net force experienced by charge Q at centroid of the triangle.

Calculation:

Distance of centroid from vertex :

$d = \dfrac{2}{3} \times \bigg \{height \bigg \}$

$= > d = \dfrac{2}{3} \times \bigg \{ \dfrac{ \sqrt{3} }{2}l \bigg \}$

$= > d = \dfrac{l}{ \sqrt{3} }$

Now force on Q due to q1 :

$\therefore \: F1 = \dfrac{kQq}{ {d}^{2} }$

$= > \: F1 = \dfrac{kQq}{ {( \frac{l}{ \sqrt{3} }) }^{2} }$

$= > \: F1 = \dfrac{3kQq}{ {l}^{2} }$

Similarly , force on Q due to q2:

$\therefore \: F2 = \dfrac{3kQq}{ {l}^{2} }$

Similarly , force on Q due to q3:

$\therefore \: F3 = \dfrac{3kQq}{ {l}^{2} }$

Now , F1 , F2 and F3 are three equal vectors directed at an angle of 120° to each other. (Refer to the diagram)

Now , whenever three equal vectors are present at an angle of 120 degrees to each other the net resultant of the vector becomes zero.

So , net force experienced by Q is ZERO.