- #1

- 34

- 0

But minus sine comes when we evaluate the dot product first.

How does the minus sign occur without evaluating the dot product?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Zubair Ahmad
- Start date

- #1

- 34

- 0

But minus sine comes when we evaluate the dot product first.

How does the minus sign occur without evaluating the dot product?

- #2

- 17,429

- 8,420

$$\vec{n} \cdot (\vec{D}_1-\vec{D}_2)=\sigma.$$

Here ##\vec{D}_1## (##\vec{D}_2##) denotes the value of ##\vec{D}## when approaching the point on the boundary surface under investigation from side 1 (side 2) of the boundary surface. The minus in the above equation comes from the fact that the surface normal vector of the pill box parallel to the boundary at side 2 is just ##-\vec{n}##, where ##\vec{n}## is the direction of the surface normal vector at side 1 (see again Fig. 7.48, where in my notation ##\vec{a}=a \vec{n}##).

- #3

- 34

- 0

Still confusing!

- #4

- 17,429

- 8,420

Why? Where is your problem?

- #5

- 34

- 0

I'm saying that negative sign would come due to dot product.so it can't be written before it.

- #6

- 17,429

- 8,420

$$a \vec{n} \cdot \vec{D}_1 + a (-\vec{n}) \cdot \vec{D}_2 = a \vec{n} \cdot (\vec{D}_1-\vec{D}_2).$$

Look again at the figure in the book. It's really quite ovious.

Share: