Question

BACK NEXT In the figure, the particles have charges q1 42 520 nc and 9394-94 nc, and distance components of the net electrostatic force on particle 3 4.1 m . Wwhat are the (o) - and (b) y Units N (a) Number Units N (b) Number SHOW HINT 38 PM

Answer 1

a )

given

q₁ = - q₂ = 520 n C

q₃ = - q₄ = 94 n C

a = 4.1 cm

= 0.041 m

x - component of particle 3 is F_x = F_1x + F_2x + F_3x

due to charge q₁ the horizontal force is zero

F_3x = F_1x + F_2x + F_4x

= 0 + F₂ cos45 + F_4x

we have basic equation F = k q₁ q₂ / r²

= k q₂ q₃ / r₂₃² x cos45 + k q₄ q₃ / r₃₄²

here r₂₃ = ( 0.041² + 0.041² )^1/2

= 0.0578 m

F_3x = k q₃ ( q₂ / r₂₃² x cos45 + q₄ / r₃₄² )

= 9 x 10⁹ x 94 x 10^-9 ( 520 x 10^-9 x cos45/ 0.058² + 94 x 10^-9 / 0.041² )

F_3x = 846 x 1.6522 x 10^-4

F_3x = 0.1397 N

b )

same way

F_3y = k q₃ ( q₂ / r₂₃² x sin45 - q₁ / r₁₃² )

= 9 x 10⁹ x 94 x 10^-9 ( 520 x 10^-9 x sin45/ 0.058² - 520 x 10^-9 / 0.041² )

= - 846 x 2.000354 x 10^-4

F_3y = - 0.16923 N