Question

Two uniformly charged, infinite, nonconducting planes are parallel to a yz plane and positioned at x " #50 cm and x " )50 cm. The charge densities on the planes are #50nC/m2 and )25 nC/m2 , respectively. What is the magnitude of the potential difference between the origin and the point on the x axis at x " )80 cm? (Hint: Use Gauss’ law.)

Answer 1

Two uniformly charged, infinite, nonconducting planes are parallel to a yz plane and positioned at x = -50 cm and x = +50 cm. The charge densities on the planes are -50 nC/m² and +25 nC/m², respectively. What is the magnitude of the potential difference between the origin and the point on the x axis at x = +80 cm? (Hint:Use Gauss’ law.)

Answer:

ΔV = 2520 V

Step-by-step explanation:

We are given;

Charge density 1; σ1 = -50 nC/m² = -50 × 10^(-9) nC/m²

Charge density 2; σ2 = +25 nC/m² = +25 × 10^(-9) nC/m²

Now, formula for electric field strength is;

E = -(½ε)(|σ1| ± |σ2|)

ε is vacuum permittivity with a constant value as 8.85 × 10^(−12) C²/N.m²

|σ1| and |σ2| means we are taking the absolute values and would therefore not use the negative sign attached to them.

Now, in between x = 0 cm(0 m) and 50 cm (0.5 m), electric field generated by both charge densities would be in the same direction and thus;

E_in = -(½ε)(|σ1| + |σ2|)

E_in = -(½ × 8.85 × 10^(−12)) × [(50 × 10^(-9)) + (25 × 10^(-9))]

E_in = -4200 V/m

In contrast, when x ≥ 50 cm (0.5 m), electric field generated by both charge densities would be in opposite directions and thus;

E_out = -(½ε)(|σ1| - |σ2|)

E_out = -(½ × 8.85 × 10^(−12)) × [(50 × 10^(-9)) - (25 × 10^(-9))]

E_out = 1400 V/m

The magnitude of the potential difference from the origin to x = 80 cm(0.8 m) is calculated as attached;

From the attachment, we see that;

ΔV = 2520 V