Question

The charges and coordinates of two charged particles held fixed in an xy plane are q1 = 3 39 μ x1 = 5 84 m. Уі О.216 c and q magnitude and b) direction with respect to +x-axis in the range -180。180° of the electrostatic force on particle 2 due to particle charge q3-6.99 μС be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero? 3.47 C X 233 m. 2.15 cr Find he . At what e and d coordinates should a third particle of (a) Number (b) Number (c) Number (d) Number Units Units Units Units

Answer 1

a )

given

q₁ = 3.39 uC

x₁ = 5.84 cm

= 0.0584 m

y₁ = 0.216 cm

= 0.00216 m

q₂ = - 3.47 uC

x₂ = - 2.83 cm

= - 0.0283 m

y₂ = 2.15 cm

= 0.0215 m

a )

r₁₂ = ( ( x₂ - x₁ )² + (y₂ - y₁ )² )^1/2

= ( -0.0283 - 0.0584 )² + ( 0.0215 - 0.00216 )² )^1/2

r₁₂ = 0.088 m

F = k q₁ q₂ / r₁₂²

= 9 x 10⁹ x 3.39 x 10^-6 x 3.47 x 10^-6 / 0.088²

F = 13.67 N

b )

angle = tan^-1 ( 2.15 - 0.216 / -2.83 - 5.84 )

= - 12.57^o

c )

r = r₁₂ ( q₃ / q₁ )^1/2

= 0.088 x ( 6.99 / 3.39 )^1/2

= 0.1263 m

x₃ = x₂ - r cos(- 12.57^o )

= - 0.0283 - 0.1263 x 0.967

= - 0.1504 m

d )

y₃ = y₂ - r sin(- 12.57^o )

= - 0.0215 + 0.1263 x 0.217

= 5.907 x 10^-3 m