Question

A -10.0 nC point charge and a +20.0 nC point charge are 15.0 cm apart on the x-axis.

Part A)

What is the electric potential at the point on the x-axis where the electric field is zero? Answer in V

Part B)

What is the magnitude of the electric field at the point on the x-axis, between the charges, where the electric potential is zero? Answer in N/C

Answer 1

A)

Let the point P where the electric field is zero be at a distance r from -10nC on the x axis. The distance between the charges is 15 cm or 0.15 m

Electric field due to -10 nC

E₁ = K(-10x10^-9)/r²

Electric field due to 20 nC

E₂ = K(20x10^-9)/(0.15 – r)²

Total field E = E₁ + E₂

E = K(-10x10^-9)/r² + K(20x10^-9)/(0.15 – r)²

0 = -K(10x10^-9)/r² + K(20x10^-9)/(0.15 – r)²

K(10x10^-9)/r² = K(20x10^-9)/(0.15 – r)²

1/r² = 2 / (0.15 – r)²

(0.15 – r)² = 2r²

0.0225 + r² – 0.3r = 2r²

0 = r² + 0.3 r -0.0225

r = 0.0621 m

r = 6.21 cm

so potential at P due to -10 nC

V₁ = k(-10nC)/r

V₁ = 9 x 10⁹ x (-10) x 10^-9/ 0.0621

V₁ = - 1449.2 V

potential at P due to 20 nC

V₂ = k(20nC)/0.15-r

V₂ = 9 x 10⁹ x (20) x 10^-9/ 0.15-0.0621

V₂ = 2047.8 V

V = v₁ + v₂

V = -1449.2 +2047.8

V = 598.6 V

B)

The potential at point P due to -10nC

v₁ = k x (-10x 10^-9) / r

The potential at point P due to 20nC

v₂ = k x 20x 10^-9/(0.15-r)

Total potential

V = v₁ + v₂

0 = k x (-10x 10^-9) / r + k x 20x 10^-9/(0.15-r)

10k x 10^-9/r = 20kx10^-9/(0.15-r)

0.15-r = 2r

3r = 0.15 m

r = 0.05 m

The field at point P due to -10nC

E₁ = k(-10nC)/r²

E₁ = 9 x 10⁹ x (-10) x 10^-9 / 0.05²

E₁ = - 36000 N/C

The field at point P due to 20nC

E₂ = k(20nC)/(0.15-r)²

E₂ = 9 x 10⁹ x (20) x 10^-9 / (0.15-0.05)²

E₂ = 18000 N/C

Total Field

E = E₁ + E₂

E = - 18000 N/C

the negative sign indicates that the field is acting along the –x axis

The magnitude of the field E = 18000 N/C