Question

Two charges of +200 uC and -200 uC are placed at the corners B and C of an equilateral triangle ABC of side 0.1 m. The force on a charge of 5uC placed A is
1) 1800 N
2)1200√3N
3) 600√3N
4) 900N​

Answer 1

Answer:

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Answer 2

Given info : two charges +200uC and -200uC are placed at the corner of B and C of an equilateral ∆ABC of side length 0.1 m.

To find : the force on a charge of 5 uC placed at A is ...

solution : force experienced by A due to B , F₁ = k(200uC)(5uC)/(0.1)²

= (9 × 10^9 × 200 × 10^-12 × 5 × 10^-6)/10^-2

= 900 N (along BA)

Force experienced by A due to C, F₂ = k(200uC)(5uC)/(0.1)²

= 900 N ( along AC)

angle between force vectors F₁ and F₂ is 120° [shown in figure. ]

Resultant force, F = √{F₁² + F₂² + 2F₁F₂cos120°}

= √{900² + 900² + 2 (900)(900)(-1/2)}

= 900 N

Therefore the force acting on the charge placed at A is 900 N. I.e., option (4)