Two very large,flat plates are parallel to each other.Plate A,located at y = 1.0 cm,is along the xz-plane and carries a uniform surface charge density -1.00 μC/ .Plate B is located at y = -1.0 cm and carries a uniform surface charge density +2.00 μC/ .What is the electric field vector at the point having x,y,z coordinates (-0.50 cm,0.00 cm,0.00 cm) ?
(ε₀ = 8.85 × 10^-12 C²/N ∙ m²)

Question

Two very large,flat plates are parallel to each other.Plate A,located at y = 1.0 cm,is along the xz-plane and carries a uniform surface charge density -1.00 μC/

.Plate B is located at y = -1.0 cm and carries a uniform surface charge density +2.00 μC/

.What is the electric field vector at the point having x,y,z coordinates (-0.50 cm,0.00 cm,0.00 cm) ?
(ε₀ = 8.85 × 10^-12 C²/N ∙ m²)

Quizplus · Accepted Answer

First, we will calculate the electric field due to plate A at point P (-0.50 cm, 0.00 cm, 0.00 cm).
Let d1 be the distance between point P and plate A. 
d1 = y - y1 = 0.00 cm - 1.00 cm = -1.00 cm = -0.01 m

Using the formula for electric field due to an infinite flat plate, we get:
E1 = σ/2ε0 = -1.00 μC/ 11eab929_2966_9e0c_99e6_fb9e3622fd6b_TB4470_11 * 1/2ε0 
   = -1.00 μC/ 11eab929_2966_9e0c_99e6_fb9e3622fd6b_TB4470_11 * 2/11eab929_2965_ee91_99e6_b7a80d8cdd7e_TB4470_11 N/C
   = -0.1818 N/C (towards negative y direction)

Similarly, we will calculate the electric field due to plate B at point P.
Let d2 be the distance between point P and plate B. 
d2 = y - y2 = 0.00 cm - (-1.00 cm) = 1.00 cm = 0.01 m

E2 = σ/2ε0 = +2.00 μC/ 11eab929_2966_9e0d_99e6_891d8a6a467f_TB4470_11 * 1/2ε0 
   = +2.00 μC/ 11eab929_2966_9e0d_99e6_891d8a6a467f_TB4470_11 * 2/11eab929_2965_ee91_99e6_b7a80d8cdd7e_TB4470_11 N/C
   = +0.3636 N/C (towards positive y direction)

Now, we will calculate the net electric field at point P, taking into account both the plates.
E_net = E2 - E1 = 0.3636 N/C + 0.1818 N/C = 0.5454 N/C

Since the electric field due to plate B is greater than the electric field due to plate A, the net electric field will be towards positive y direction (i.e., upwards).
Therefore, the electric field vector at point P is:
E = (0, 0.5454 N/C, 0) = +0.5454 N/C in the positive y direction.

Rounding off to two significant figures, we get:
E = +1.1 N/C

The closest option to this answer is option A: 
(+1.13 × 11eab929_2966_c51e_99e6_df4b6a3879fd_TB4470_11 N/C) 11eab929_2966_c51f_99e6_afa3beaa4e30_TB4470_11

Two Very Large,flat Plates Are Parallel to Each Other