Question

Three identical point charges of Q coul are placed at the vertices of an equilateral triangle, 10 cm apart. Calculate the force on each charge

Answer 1

Answer:

200k newtons

Explanation:

assume all charges are positive unit charges

F1 = Kq1q2/d² = k/0.01 = 100k

F2 = 100k

resultant = 200k√1+cos 60° = 400sin30° = 200k newtons

Answer 2

Three identical point charges of Q Coul are placed at the vertices of an equilateral triangle

Explanation:

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The distance between charges , d= 10 cm

We know that ,the force between to point charge

$F=K\dfrac{q_1q_2}{d^2}$

K=9 x 10⁹

From the diagram

$F_1=F_2=F$

$F=9\times 10^9\dfrac{Q.Q}{0.1^2}$

$F=9 Q^2\times 10^{11}\ N$

The angle between F₁ and F₂ is 60⁰ , therefore the resultant force R

$R^2=F_1^2+F_2^2+2F_1F_2cos\theta$

$R^2=F^2+F^2+2F^2cos60^{\circ}$

$R^2=F^2+F^2+2F^2\times \dfrac{1}{2}$

$R=\sqrt{3}\ F$

$R=\sqrt{3} \times 9 Q^2\times 10^{11}\ N$