Calculate Electric Flux from Point Charge to Plate:

In summary, the problem asks for the electric flux from a point charge to a flat plate located 5 cm away on the x axis. The solution can be found using surface integrals in both Cartesian and spherical coordinates. In Cartesian coordinates, the integral can be written as the product of the charge and the distance between the charge and the point on the plate, divided by the cube of the distance. In spherical coordinates, the integral can be simplified using trigonometric identities. The integral of sine raised to an odd power is straightforward, but for cosine raised to an even power, it can be converted into a sine function.
  • #1
Faiq
348
16

Homework Statement


A ##10 cm## (on y axis) by ##10 cm## (on z axis) flat plate is located ##5 cm## away (on x axis) from a point charge ##q##. Calculate the electric flux from the point charge to the plate.

Can somebody solve it using surface integral using both spherical and cartesian coordinates. I did solve it using symmetry but I don't know how to solve it using integrals?

Cartesian
$$ \oint_s \vec{E}^\ \cdot \vec{n}^\ dS= \iint \frac{\lambda}{4\pi \varepsilon_0} \frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^3}\cdot \vec{n}^\ dS = \iint \frac{\lambda}{4\pi \varepsilon_0} \frac{x-x_{0}}{((x-x_0)^2+y^2+z^2)^\frac{3}{2}} dS = \iint \frac{\lambda}{4\pi \varepsilon_0} \frac{x-x_{0}}{((x-x_0)^2+y^2+z^2)^\frac{3}{2}} dydz$$
where ##x_0## is ##5cm##

Spherical
$$ dydz = \frac{\partial(y,z)}{\partial (\phi,\theta)}d\phi d\theta = -\sin^2\phi \cos \theta ~d\phi d\theta$$
$$ \oint_s \vec{E}^\ \cdot \vec{n}^\ dS= \iint \frac{\lambda}{4\pi \varepsilon_0} \frac{ dS}{|\mathbf{r}|^2}i\cdot \vec{n}^\ = \iint \frac{\lambda}{4\pi \varepsilon_0} \frac{ dS}{|\mathbf{r}|^2}(\cos \theta \sin \phi ) = -\iint \frac{\lambda}{4\pi \varepsilon_0} \frac{ d\phi d\theta}{|\mathbf{r}|^2}(\cos^2 \theta \sin^3 \phi ) = \iint \frac{\lambda}{4\pi \varepsilon_0} \frac{ dS}{|\mathbf{r}|^2}(\cos \theta \sin \phi ) = -\iint \frac{\lambda}{4\pi \varepsilon_0} \frac{ d\phi d\theta}{x_0^2}(\cos^4 \theta \sin^5 \phi ) $$
 
Last edited:
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  • #2
##\int \sin^5(\phi).d\phi## is fairly straightforward.
What would you do with ##\int \cos^2##?

(Is the plate centred on the x axis?)
 
  • #3
Convert into a sine function?
Yes
 
Last edited:
  • #4
Faiq said:
Convert into a sine function?
Yes
No. Cos(2x)=?
 

1. How do you calculate the electric flux from a point charge to a plate?

To calculate the electric flux from a point charge to a plate, you will need to use the formula F = q/ε₀A, where F is the electric flux, q is the charge of the point charge, ε₀ is the permittivity of free space, and A is the area of the plate.

2. What is the unit of electric flux?

The unit of electric flux is Nm²/C, which is equivalent to volts per meter (V/m).

3. How does the distance between the point charge and the plate affect the electric flux?

The electric flux is inversely proportional to the distance between the point charge and the plate. This means that as the distance increases, the electric flux decreases, and vice versa.

4. Can the electric flux be negative?

Yes, the electric flux can be negative. This occurs when the electric field and the area vector are in opposite directions.

5. What factors can affect the electric flux from a point charge to a plate?

The electric flux from a point charge to a plate can be affected by the magnitude of the charge, the distance between the charge and the plate, and the orientation of the plate with respect to the electric field. The permittivity of free space also plays a role in determining the electric flux.

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