# Find the Angle Theta for E Field in Glass Below Boundary | E Field Problem

• austinmw89
In summary, the E field in the glass just below the boundary is equal to 3 V/m. The angle theta for the field is 61.9°.
austinmw89
Problem statement:
The E field measured just above a glass plate is equal to 2 V/m in magnitude and is direct at 45° away from the boundary, as shown in the figure. The magnitude of the E field measured just below the boundary is equal to 3 V/m. Find the angle theta for the field in the glass just below the boundary.

(sorry I don't know why uploading the image flipped it upside down)

I have the solution from the manual for this problem, but I still don't understand it. I'd really appreciate if someone could explain the beginning set-up of the solution:

E_1tan = E_2tan -> E_1cos(theta)=E_2cos(45°) -> cos(theta) = (E_2/E_1)*(sqrt(2)/2) -> theta = 61.9°.

I can follow the math from the second step, but my geometry is pretty rusty and I don't understand how they start with the tangents and also how they go from tangents to cosines.

Last edited:
"tan" stands for tangential, the tangential component of the E field to the boundary. You simply have to use the EM field boundary conditions. You basically have two options of which boundary conditions (there are 4 of them) will be helpful for this problem. Since you are given the E fields, the relevant boundary conditions are those involving the components of E field:
$$(i) \hspace{2mm} \epsilon_1 E_1^{\perp} - \epsilon_2 E_2^{\perp} = \sigma_f$$
$$(ii) \hspace{2mm} \mathbf{E_1}^{\parallel} = \mathbf{E_2}^{\parallel}$$
Now since you are not given the permittivities ##\epsilon## and also the free charge distribution ##\sigma_f## , the best candidate to choose is the second one.

Last edited:
austinmw89
So if the boundary condition for the E field is [normal X (E_1 - E_2) = 0] I get [normal X E_1 = normal X E_2]. How do I go from this to tangent?

The normal and tangential boundary conditions are independent of each other, see my edited previous comment to see why the conditions for the normal components should not be used in this case.

The tangential components of the electric field must be continuous across any interface. So the component of the electric field inside the glass tangential to the interface E_1 Cos(\theta) must be equal to the tangential component of the electric field in air E_2 Cos(\Pi/4). Equating the two and solving for the angle \theta gives your result.

Reason: In order for the field (electric or magnetic) to be completely specified in both glass and air (or any material for that matter) the tangential field components must be continuous across any interface. This is know as the Uniqueness Theorem.

austinmw89
Thank you both

## 1. What is the equation for finding the angle theta for E field in glass below a boundary?

The equation for finding the angle theta is: sin(theta 1)/sin(theta 2) = n2/n1, where n1 and n2 are the refractive indices of the two materials.

## 2. How do I know which material is n1 and which is n2?

The material with the lower refractive index is typically designated as n1 and the material with the higher refractive index is designated as n2. However, this may vary depending on the specific problem.

## 3. Is there a specific unit for the angle theta?

The angle theta is typically measured in radians, but it can also be measured in degrees. It is important to check the units in the given problem and convert if necessary.

## 4. Can I use this equation for any type of boundary between two materials?

This equation is specifically for finding the angle theta for an electric field in glass below a boundary. It may not be applicable to other types of boundaries or materials. It is important to check the given problem and use the appropriate equation.

## 5. What is the significance of finding the angle theta for the electric field in glass below a boundary?

The angle theta represents the direction of the electric field as it passes through the boundary between two materials. It helps us understand how the electric field is affected by the change in material, and can be used to calculate other properties such as the intensity of the electric field.

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