Question

three parallel sheets of charge, large enough to be treated as infinite sheets, are perpendicular to the x - axis. sheet a has surface charge density \\(\sigma_a = + 8.00\\ nc/m^2\\). sheet b is 4.00 cm to the right of sheet a and has surface charge density \\(\sigma_b = - 5.00\\ nc/m^2\\). sheet c is 4.00 cm to the right of sheet b, so is 8.00 cm to the right of sheet a, and has surface charge density \\(\sigma_c = + 6.00\\ nc/m^2\\). \\(\boldsymbol{\text{part a}}\\) what are the magnitude and direction of the resultant electric field at a point that is midway between sheets b and c, or 2.00 cm from each of these two sheets? express your answer in newtons per coulomb. enter positive value if the field points in +x - direction and negative value if the field points in -x - direction. \\(e_x = \\) \\(\boxed{\space}\\) n/c

Explanation:

Step1: Recall field of infinite sheet

The electric field due to an infinite charged sheet is $E = \frac{\sigma}{2\epsilon_0}$, where $\epsilon_0 = 8.85 \times 10^{-12} \frac{\text{C}^2}{\text{N·m}^2}$.

Step2: Define field directions

For a point midway between B and C:

• Sheet A ($\sigma_A=+8.00\ \text{nC/m}^2=8.00\times10^{-9}\ \text{C/m}^2$): Field points away, so $+x$ direction.
• Sheet B ($\sigma_B=-6.00\ \text{nC/m}^2=-6.00\times10^{-9}\ \text{C/m}^2$): Field points toward, so $+x$ direction.
• Sheet C ($\sigma_C=+6.00\ \text{nC/m}^2=6.00\times10^{-9}\ \text{C/m}^2$): Field points away, so $-x$ direction.

Step3: Calculate individual fields

$E_A = \frac{8.00\times10^{-9}}{2\times8.85\times10^{-12}} \approx 451.98\ \text{N/C}$
$E_B = \frac{|-6.00\times10^{-9}|}{2\times8.85\times10^{-12}} \approx 338.98\ \text{N/C}$
$E_C = \frac{6.00\times10^{-9}}{2\times8.85\times10^{-12}} \approx 338.98\ \text{N/C}$

Step4: Sum fields with signs

$E_x = E_A + E_B - E_C$
$E_x = 451.98 + 338.98 - 338.98 = 451.98\ \text{N/C}$

Answer:

$452$ N/C (rounded to 3 significant figures, positive value indicates +x direction)