Physics, 21.02.2020 17:10 saltytaetae

An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge density σ₁ = 0.44 μC/m². Another infinite sheet of charge with uniform charge density σ₂ = -0.66 μC/m² is located at x = c = 35 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i. e., between x = 15.5 cm and x = 19.5 cm).
1) What is Ex(P), the x-component of the electric field at point P, located at (x, y) = (7.75 cm, 0)? N/C
2) What is , the charge density on the surface of the conducting slab at x = 15.5 cm μC/m²
3) What is V(R) - V(P), the potential difference between point P and point R, located at (x, y) = (7.75 cm, -19.5 cm)? V
4) What is V(S) - V(P), the potential difference between point P and point S, located at (x, y) = (27.25 cm, -19.5 cm)? V
5) What is Ex(T), the x-component of the electric field at point T, located at (x, y) = (42.75 cm, -19.5 cm)? N/C

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Answer from: anrs14

1) $E=6.21\cdot 10^4 V/m$

2) $\sigma_a = -1.10\mu C/m^2$

3) V(R)-V(P)=0

4) V(S)-V(P)=-9625 V

5) $E=-1.24\cdot 10^4 V/m$

Explanation:

The electric field outside an infinite sheet of charge is

$E=\frac{\sigma}{2\epsilon_0}$

where

$\sigma$ is the surface charge density

$\epsilon_0$ is the vacuum permittivity

And it is perpendicular to the sheet (outward if the

Here we have:

- An infinite sheet of charge located at x = 0, with uniform charge density $\sigma_1 = 0.44\mu C/m^2 = 0.44\cdot 10^{-6} C/m^2$

- Another infinite sheet of charge located at x = 35 cm, with charge density $\sigma_2 = -0.66\mu C/m^2 = -0.66\cdot 10^{-6} C/m^2$

- And an infinite uncharged conducting slab located between x = 15.5 cm and x = 19.5 m

Since the slab is globally uncharged, the electric field produced by the slab outside the slab is zero. So, the electric field at

x = 7.75 cm

Is simply equal to the vector sum of the fields generated by the two sheets; at x = 7.75 cm, both fields point to the right (because sheet 1 is positive while sheet 2 is negative), therefore:

$E=\frac{\sigma_1}{2\epsilon_0}-\frac{\sigma_2}{2\epsilon_0}=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(0.44\cdot 10^{-6}-(-0.66\cdot 10^{-6}))=6.21\cdot 10^4 V/m$

And the direction is to the right.

2) The slab is conductive, so the charges in the slab redistribute such that the electric field inside the slab is zero.

Therefore, the slab can be seen as consisting of two infinite sheets, of opposite charge $+\sigma_a$ and $-\sigma_a$ .

The electric field due to the two outer infinite sheets is the one calculated in part A:

$E_{sheets}=6.21\cdot 10^4 V/m$ to the right

The electric field generated by the two charge densities on the outer surfaces of the slab is:

$E_{slab}=\frac{1}{2\epsilon_0}(\sigma_a-(-\sigma_a))=\frac{\sigma_a}{\epsilon_0}$

As we said, the net field inside the slab must be zero, so:

$E=E_{sheets}-E_{slabs}=0\\E_{slabs}=\frac{\sigma_a}{2\epsilon_0}=6.21\cdot 10^4\\\sigma_a=2\epsilon_0 (6.21\cdot 10^4)=1.10\cdot 10^{-6} C/m^2=1.10\mu C/m^2$

And sincde the field produced by the sheets is towards the right, the field produced by the slab must be to the left: so, left surface (x=15.5 cm) must be negatively charged, while the right surface (x=19.5 cm) must be positively charged. So the charge density at x = 15.5 cm is

$\sigma_a = -1.10\mu C/m^2$

Here all the electric fields are perpendicular to the sheets: this means that they do not have components in the vertical direction, so we can ignore any difference in the y-coordinate and consider only the differences in the x-coordinate.

Here the two points are:

P = x = 7.75 cm

R = x = 7.75 cm

The two points have same x-coordinate. This also means that the two points have exactly same electric field, since it has no components along the vertical direction: therefore, the potential difference between the two points must be zero,

V(R)-V(P)=0

As before, we can totally ignore any difference in the y-coordinates since it is irrelevant, so the two points in this case are

P = x = 7.75 cm

S = x = 27.25 cm

Point P is located between sheet 1 and the conductive slab, while point S is located between the slab and sheet 2.

FIrst of all, we observe that the field is uniform between the two sheets (and zero in the region inside the slab). For a uniform field, the potential difference between two points is given by

$\Delta V = E d$

where E is the magnitude of the electric field and d the distance between the two points.

Here d is the distance between point P and point S, however we have to subtract from it the region of space where there is no electric field (between 15.5 cm and 19.5 cm), so:

d=(27.25 cm-7.75 cm)-(19.5 cm-15.5 cm)=15.5 cm=0.155 m

Therefore, the potential difference is

$V(S)-V(P)=(6.21\cdot 10^4)(0.155)=9625 V$

However, the electric field is to the right, this means that point P is at higher potential than point S: so,

V(S)-V(P)=-9625 V

Again, we only consider the x-coordinate of the point T, which is

x = 42.75 cm

Here we are on the right of sheet 2. This means that at this point:

- The field produced by sheet 1 points to the right

- The field produced by sheet 2 points to the left

This means that the two fields point in opposite directions, so the net field will be the difference between the two fields.

Therefore, we have:

$E(T)=\frac{1}{2\epsilon_0}(\sigma_1 + \sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(0.44\cdot 10^{-6}-0.66\cdot 10^{-6})=-1.24\cdot 10^4 V/m$

And the negative sign means the field is to the left.

Answer from: Quest

Iwould sure like to answer your question but i honestly just don’t know

Answer from: Quest

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr

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