Question

Two large, thin metal plates are placed parallel and
close to each other. Their inner faces have equal and
opposite charges with a surface charge density of
17.0 x 10^-12 C m^-2. What are the magnitudes of
electric fields to the left of the plates, to the right of
the plates and in between the plates.​

Answer 1

Answer:

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A and B are two parallel plates close to each other. Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II.

Charge density of plate A, σ = 17.0 × 10^−22 C/m²

Charge density of plate B, σ = −17.0 × 10^−22 C/m²

In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the respective plates.

Electric field E in region II is given by the relation,

$E = \frac{σ}{∈0</p><p>}$

Where,

∈0 = Permittivity of free space = 8.854 × 10^−12 N^−1C² m^−2

∴

$E = \frac{17.0×10 {}^{ - 22} </p><p>}{8.854 \times 10 {}^{ - 12} }$

= 1.92 × 10^−10 N/C

Therefore, electric field between the plates is 1.92 × 10−10 N/C.

Answer 2

Answer:

A and B are two parallel plates close to each other. Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II.

Charge density of plate A, σ = 17.0 × 10^−22 C/m²

Charge density of plate B, σ = −17.0 × 10^−22 C/m²

In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the respective plates.

Electric field E in region II is given by the relation,

E = \frac{σ}{∈0 }E=

∈0

σ

Where,

∈0 = Permittivity of free space = 8.854 × 10^−12 N^−1C² m^−2

∴

E = \frac{17.0×10 {}^{ - 22} }{8.854 \times 10 {}^{ - 12} }E=

8.854×10

−12

17.0×10

−22

= 1.92 × 10^−10 N/C

Therefore, electric field between the plates is 1.92 × 10−10 N/C.