Question

three particles have charges 20 new coulomb they are fixed in equilateral triangle of side 0.5m .The force on each particle

Answer 1

Answer:

$\bigstar{\bold{Force=25\:N}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Three particles having charges = 20 μ C = 20 × 10⁻⁶ C
• Side of equilateral triangle = 0.5 m

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Force on each particle

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ The force between between two charges is given by the formula,

   $F=\dfrac{1}{4 \pi E_0 } \dfrac{q_1q_2}{r^{2} }$

→ Here q₁ = q₂ = 20 × 10⁻⁶

  r = 0.5 m

 $\dfrac{1}{4\pi E_0 } = 9\times 10^{9}$

→ Substituting the datas we get

$F=\dfrac{9\times 10^{9}\times (20\times 10^{-6})^{2} }{(0.5)^{2} }$

$F=\dfrac{9\times 10^{9}\times 400\times 10^{-12} }{0.25}$

$F=\dfrac{3600\times 10^{-3} }{0.25}$

F =  14.4 N

→ The resultant force is given by

$R=\sqrt{F^{2}+F^{2}+2F^{2}cos\:60$

$R=\sqrt{(14.4)^{2}+(14.4)^{2}+2\times (14.4)^{2}\times \dfrac{1}{2} }$

R = √622.08

R = 25 N

→ Here force acting on each charge is same since the magnitude of charges and distance between them are similar.

→∴ Force on each charge = 25 N

$\boxed{\bold{Force=25\:N}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ Coulumb's law state that the force of interaction between two point charges is  directly proportional to the product of magnitude of the charges and inversely proportional to the square of distance betweenn them.

$F=\dfrac{1}{4 \pi E_0 } \dfrac{q_1q_2}{r^{2} }$