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.)

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.)

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

he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write   -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get

and solving the given expression :

3x(2x-1)

since solving the expression using tiles   and solving the expression algebraically we get the same answer

thus   we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".

idek

step-by-step explanation: