Any equation in the form
Is a line. So, we only need to sample two points, and then connect them.
If we choose, for example, x=0 and x=3, we have
So, when x=0 then y=0, i.e. the line passes through the origin (0,0).
Similarly, when x=3 then y=2, i.e. the line passes through the point (3,2).
Connect these two points and you have the line.
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
