General Physics II

23.***. A charge of - 4.0 microCoulombs is located at the origin and a charge of - 5.0 microCoulombs is located along the y-axis at y = 2.0 m. At what point along the y-axis is the electric field zero?

E = k q/r²

r₁ = y

r₂ = (2.0 m - y)

r₂² = (2.0 m - y)²

r₂² = 4 - 4 y + y²

where y must now be measured in meters.

For point P between the two charges, 0 < y < 2.0 m, E₁ points down and E₂ points up so the total field is E_tot = E_net = E₂ - E_{1 .}

E_tot = E_net = E₂ - E₁= 0

E_tot = k q₂/r₂² - k q₁/r₁²

We have already taken care of the direction of the electric field due to the signs of the charges and the location of point P. In the following equations the q's are just the magnitudes.

E_tot = k (5 C)/(4 - 4 y + y²) - k (4 C)/y^{2
= 0}

5 /(4 - 4 y + y²) - 4 /y^{2 = 0}

5 y² /[y² (4 - 4 y + y²)] - 4 (4 - 4 y + y²) /^{[y² (4 - 4 y + y²)]
= 0}

[5 y² - 4 (4 - 4 y + y²)] /^{[y²
(4 - 4 y + y²)] = 0}

[5 y² - 4 (4 - 4 y + y²)] ^{= 0}

5 y² - 4 (4 - 4 y + y²)^{= 0}

5 y² - 16 + 16 y - 4 y²^{= 0}

y² - 16 + 16 y ^{= 0}

y² + 16 y - 16 ^{= 0}

We can now solve this using the quadradic equation; the solutions are

y = - 8.06 m or y = + 0.94 m

Of course, the PHYSICS of the situation requires that

y = 0.94 m

Ch23 Homework Solutions