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.