Question

Please see question below. Please double check all work and units/unit conversions. Show all work. Thank you!

2a) A charge of -2.8 micro-coulombs lies at the origin and another charge of +9.5 micro- coulombs lies at x--14 cm, y +4 cm. At a certain point, the total electric field from these two charges is zero. What is the y coordinate of this point in cm? Include a negative sign if negative. 2b) A charge of +8 micro-coulombs is placed at x-0 cm, y +17 cm, a charge of -5 micro- coulombs is placed at x-0 cm, y -24 cm, a charge of +9 micro-coulombs is placed at x -7 cm, y 0 cm, and an unknown charge electric field at the origin is 298 degrees counter-clockwise from the positive x axis. What is the unknown charge in micro-coulombs? If negative, include a negative sign in your answer is placed! at x +25 cm, y = 0 cm. The angle of the total

Answer 1

2a)

consider the triangle AOB

using Pythagorean theorem

r = distance between the two charges = AB = sqrt(OA² + OB²) = sqrt(0.04² + 0.14²) = 0.15 m

$\theta$ = tan^-1(OA/OB) = tan^-1(0.04/0.14) = 15.94 deg

consider point C

for the electric field at C to be zero

E₁ = E₂

k q₁/d² = k q₂/(r + d)²

q₁/d² = q₂/(r + d)²

(2.8 x 10^-6)/d² = (9.5 x 10^-6)/(0.15 + d)²

d = 0.18 m

y-coordinate : - d Sin$\theta$ = - (0.18) Sin15.94 = - 0.0494 m = - 4.9 cm

2 b)

when q₄ is positive

Net electric field at the origin along X-direction is given as

E_x = E₃ - E₄ = k q₃/r₃² - k q₄/r₄² = (9 x 10⁹) (9 x 10^-6)/(0.07)² - (9 x 10⁹) q₄/(0.25)²

Net electric field at the origin along Y-direction is given as

E_y = E₁ + E₂ = k q₁/r₁² + k q₂/r₂² = (9 x 10⁹) (8 x 10^-6)/(0.17)² + (9 x 10⁹) (5 x 10^-6)/(0.24)² = 3.3 x 10⁶

Given that :

tan$\theta$ = E_y /E_x

tan62 = ( 3.3 x 10⁶)/E_x

E_x = 1.8 x 10⁶ N/C

when q₄ is positive

E_x = E₃ - E₄ = k q₃/r₃² - k q₄/r₄² = (9 x 10⁹) (9 x 10^-6)/(0.07)² - (9 x 10⁹) q₄/(0.25)²

1.8 x 10⁶ = (9 x 10⁹) (9 x 10^-6)/(0.07)² - (9 x 10⁹) q₄/(0.25)²

q₄ = 102.3 x 10^-6 C

when q₄ is negative :

E_x = E₃ - E₄ = k q₃/r₃² + k q₄/r₄² = (9 x 10⁹) (9 x 10^-6)/(0.07)² + (9 x 10⁹) q₄/(0.25)²

1.8 x 10⁶ = (9 x 10⁹) (9 x 10^-6)/(0.07)² + (9 x 10⁹) q₄/(0.25)²

q₄ = - 102.3 x 10^-6 C