Hi. In the given problem, these are the two large copper plates. This is first copper plate than in parallel. With this. There is another copper plate which are having the charge is spread over them as positive and negative. 4.0. Cool. Um, 4 m squared. This is not a charge. The sister These are the charge densities. Sigma, the surface charge Density is over. Both off. These copper plates are given us having the same magnitude but opposite nature. The first plate is having a positive charge and the second plate is having a negative charge. So the electric field between the players will be like this away from the positive charge and two words the negative charge. So the electric field between these plates will be given by cause the serum using the concept, of course. The serum it will be given by Sigma by Absolutely not. So here it becomes 4.0 Coolum for meter square. The magnet, you know, Charge density divided by permitted Vitti having a value off 8.854 into 10 days. Par minus 12 Coolum Square, Part Newton Meter square, which comes out Toby, 4.52 in tow. 10 days for 11 Newton per column. Now, in the first part of the problem, we have to find the electric flux passing through a rectangular area, which is kept just between these plates. And in the first case, this is skeptic Carol toe the place. So the electric plugs passing through this rectangular area, probably given by e dot e or we can see this is E a cost. Tita. We're theater is the angle between electric field vector and area vector. And, as we know, area Vector is given by an arrow drawn perpendicular to the area. So here we can consider this to be a bar, and this is a bar already so we can say in the first case angle between electric field and area. Vector is zero degree, so plugging in unknown values here, electric flags comes out Toby 4.52 in tow. 10 Dish, Part 11 Newton for column multiplied by area off this rectangular area, the angular plane whose dimensions are given as three centimeter by four centimeters so we can write it like three centimeter or 3 500 m multiplied by 4, 500 meter so Finally, this answer here comes out Toby. Five point for two into 10 days, the Part eight Newton meters Squire for Coolum, and it becomes the answer for the first part off the problem. Now, in the second part of the problem, they are saying, if the area this rectangular area is stilted at an angle of 30 degree from the parallel direction means in this case, if we re draw the diagram, these are the two copper plates kept parallel to each other. This is the direction off electric field. This was the original orientation off the rectangular area parallel to the plate, and now it is said that it is still did through an angle off 30 degree from its original position. So it is clear that the area vector will also get tilted by the same angle. So this time the angle between electric field, an area vector, will be 30 degree. So you can see in this case in second part of the problem, the flux will be given by E cost Tita, or we can say this is 4.52 in tow. 10 dish for 11 Newton per column for area. This is three by 100 into four by 100. Meet early square into cause sign off 30 degrees. So finally this answer comes out, Toby. 4.7 into 10 days to the part eight Newton meters squared for Coolum. And here it becomes the answer for the second part. Off top problem. Thank you.