Question

A 6 (mu Coulomb) point charge is located at the origin and a - 2.5 (mu Coulomb) point charge is located 1m to the right along the X-axis. Determine the point (besides infinity) at which the electric field is 0.

Please show steps and explaination. Thanks in advance!

Answer 1

Charge and their positions:

q₁ = 6 * 10^-6 C                 at       x₁ = 0 m

q₂ = -2.5 * 10^-6 C             at       x₂ = 1 m

k = 8.99 * 10⁹ N.m²/ C² is the Coulomb's force constant.

For x<0:

E = 0 = -kq₁/ x² + kq₂/ (1+x)²

8.99 * 10⁹ [-6 * 10^-6 / x² + 2.5 * 10^-6 / (1+x)²] = 0

6 / x² = 2.5 / (1+x)²

6 * (1 + 2x + x²) = 2.5 x²

3.5 x² + 12 x + 6 = 0

Solving for x, we get

x = -0.6077 m         or       x = - 2.8208 m

For 0< x< 1 m:

E = 0 = kq₁/ x² + kq₂/ (1-x)²

8.99 * 10⁹ [6 * 10^-6 / x² + 2.5 * 10^-6 / (1-x)²] = 0

6 / x² = -2.5 / (1-x)²

6 * (1 - 2x + x²) = -2.5 x²

8.5 x² - 12 x + 6 = 0

Solving for x, we find out that, x turns out to be a complex number for this range. Hence, 0 < x < 1 m, there is no point for which E = 0

For x > 1 m:

E = 0 = kq₁/ x² - kq₂/ (x - 1)²

8.99 * 10⁹ [6 * 10^-6 / x² - 2.5 * 10^-6 / (x - 1)²] = 0

6 / x² = 2.5 / (x - 1)²

6 * (1 - 2x + x²) = 2.5 x²

3.5 x² - 12 x + 6 = 0

Solving for x, we get,

x = 0.6077 m       or      x = 2.8208 m