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A nonuniform electric field is directed along the I-axis at all pointi but not with respect to y or z The axis of a cylindrical surface, parallel to the =-axis, as ...

Question

A nonuniform electric field is directed along the I-axis at all pointi but not with respect to y or z The axis of a cylindrical surface, parallel to the =-axis, as shown in the figure. The electric fields have magqitudes_of 6000 NIC and 1000 NIC respectively; and a passing through the cylindrical surface? i80 (Area Vector) Az ( Arqavochr ) 02 Pi0.20 mE20,80 m0.00 N mZic -160 N m2ic +160 N mzic 350 N - m2Ic +350 m2/cT ^ 1- 0+ | 4 AluxAlux

A nonuniform electric field is directed along the I-axis at all pointi but not with respect to y or z The axis of a cylindrical surface, parallel to the =-axis, as shown in the figure. The electric fields have magqitudes_of 6000 NIC and 1000 NIC respectively; and a passing through the cylindrical surface? i80 (Area Vector) Az ( Arqavochr ) 02 Pi 0.20 m E2 0,80 m 0.00 N mZic -160 N m2ic +160 N mzic 350 N - m2Ic +350 m2/c T ^ 1- 0+ | 4 Alux Alux



Answers

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.

Okay, so the electric foods, this problem makes an angle with 60 degrees in the vertical. So we have a plate here we have the vertical. Here we have the electric feud here on the 60 degrees angle in here. The definition off the electric flux is just the electric field. The multiplies, the surface area course sign of teeter. Just 16. This problem, the area off the plate is just a square. Therefore is we have to did that dimensions, which is D. Therefore, the area is going to be deep square. And because of that, we can say that the electric flux is going to be 3 50 They're multiplies, uh, 0.5. Sorry. Not 0.5 five times 10 to the minus two square course sign or 60 degrees. Therefore, the electric flux in this problem is just 0.4 30 eight new toes. Meters per kulaib. That's the final answer. Thanks for watching

In this problem. It has given that electric field director is equals two. Three by five energy. I kept list four by five E nod. Jacob. Okay. And where the value of the note is also mentioning the problem which is two in 2, 10 days to the Power three Newton per column. We have to find floods through a rectangular surface which is kept to paddle to the wiser plane. So if we draw the coordinate access, let us say this is X. This is why and this is dead. So we see here that if a plane is paddle to wiser texas, then it's perpendicular direction will be in the X direction. It means that if we talk about the area vector, then area vector will be close to A magnitude of the area, which is 0.2 m squared. And unit of this area vector will be or the unit normal vector to the given area vector will be in the icap direction. So it will be 2.2 I cap. Now we know that Flags. There was a fight is equals two E dot E. So there is not product between two vectors. No easy victories given in the problem. So we can put its value so it will be three upon five in order a cab bliss four upon five E. Not Jacob. And we have to take dot of these vectors with 0.2. Icap. Okay, so if we distribute this dot product so first I cap will beat. So first dot product will off I kept with the icap and the second dot product will be of icap with Jacob unit. So I kept and Jacob are the perpendicular vectors. Therefore there dot product will give zero. So the only vector that will be the only quantity that will be left after this operation will be three upon five. He not Into 0.2. And I kept out. I kept will be one. So this will be equals to three a 30.5. We can put the value of peanut which is To into 10, raised to the power three In 2, 0 two. So it will be close to ah we can Take this 10 down in the denominator. So this tin will cancel this cube with his square. So it will be close to two in 2 to four in 2, three, 12, divided by five in 2, 10 days to the power to which is equals to 1200. They were at by five and it will be equals two, 240. Therefore, flux associated with that area will be to 14. So this is the answer. Far the problem.

For this problem. On the topic of castles law, We have shown a square surface that measures 3.2 mm on each side. It is then placed in a uniform electric field which has a magnitude of 1800 newtons per column. And the feline felines at an angle of 35° with a normal to the surface. As we can see. Now, if we take the normal to be directed outward, we want to calculate the electric flux through the surface. Now we know the electric flux of the surface is given by fire, which is E dot A. Which in this case is E A. Course in peter and the angle Peter between A&E. is equal to 180° -35° Which we can see easily from the Diagram and this is 145°. And so therefore the electric flux through the surface. Fi is E a co sign theater, Which is the electric field strength, E 1800 Newtons per column times the area, which is the side squared or 3.2 times 10 to the minus three m squared Times The Co sign of 145°. This gives us the flux through the surface to be -1.5 Times 10 to the -2 newton meter squared Pakula

For this problem, on the topic of castles law, we have shown a square surface which measures 3.2 mm on each side immersed in a uniform electric field, which has a magnitude of 1800 newtons per column. And the field lines make an angle of 35° with a normal to the surface is shown. If we take the normal to be directed outward, we want to calculate the electric flux through the surface. Now the electric flux through the surface. Fi is equal to E dot the area vector A, which is E. A call sign data, and the angle Theatre between A&E is 180° minus 35°, which is 145°. and 45 degrees. And so The flux through that area five is equal to E. A call sign Theatre, which is the electric field magnitude 1800 newtons per column Times The area, 3.2 Times 10 to the -3 m squared Times The Co sign of 145°. This gives the flux through that surface to B -1.5 Times 10 to the minus two newton meter squared Pakula.


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