Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here!

# A uniform electric field of magnitude $E=435 \mathrm{N} / \mathrm{C}$ makes an angle of $\theta=65.0^{\circ}$ with a plane surface of area $A=3.50 \mathrm{m}^{2} \mathrm{as}$in Figure $\mathrm{P} 15.44$ . Find the electric flux through this surface.

## $1.38 \times 10^{3} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$

### Discussion

You must be signed in to discuss.
##### Christina K.

Rutgers, The State University of New Jersey

##### Aspen F.

University of Sheffield

##### Jared E.

University of Winnipeg

### Video Transcript

Hello. Strange. Let's start our discussion. Suppose you have a surface, Okay? No. From this. So phase the electric field are coming literally falling. The this Okay. And they're making the angle off Tito or 65 degree from that plane off this office. No, we have to find out the electric floods through the surface. So basically, the formal off electric flux is flexes a questo e a question. But the tita is the angle between the electric field in the area Vector. So from here, if electric field is like this and area factories like this, that is. And gap is in this direction. Uh, we have the anger t die here, so no, we have to find out this angle the angle between e and angle between area vector. Okay, so this anger is 90 degree and today, 65 degree. So by our triangle, some property, we will get the this angle B 25 Dickie and the electric field have the magnitude off 4 35 new timber Coolum and the area of the surface is 3.5 major square. So from these values, the flexes for 35 into 3.5 in tow course 25 degree. By solving this, we will get a value off. 1379 point it. You have done meters Well. Birth. Hola. This is our electric flux through the surface. Well, this is all for me for this video. I hope you will like the video. Thank you.

Other Schools
##### Christina K.

Rutgers, The State University of New Jersey

##### Aspen F.

University of Sheffield

##### Jared E.

University of Winnipeg