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44.4 cm4.7 cm rectangle lies the ry-plane.Yau rray want [AYiEM (Pages 864 668) _Pant AWnat I Ine eleciric Rux trough the rectangle(140i 2O0R) N/C?AZdN-m?/CSubmitEre...

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44.4 cm4.7 cm rectangle lies the ry-plane.Yau rray want [AYiEM (Pages 864 668) _Pant AWnat I Ine eleciric Rux trough the rectangle(140i 2O0R) N/C?AZdN-m?/CSubmitErevious Anskers Bcquest AnsxterIncorrect; Try Agaln; attempts remainingPartWnat the oleciric ilux tnrough the rectangle(140i 200j) N/C?N.m? /CSubmilPrevious AnswersCorrectHereprTD that when the field lines are CarallgFunacpelectric ilux tnrough this surface_ equals _

44.4 cm 4.7 cm rectangle lies the ry-plane. Yau rray want [AYiEM (Pages 864 668) _ Pant A Wnat I Ine eleciric Rux trough the rectangle (140i 2O0R) N/C? AZd N-m?/C Submit Erevious Anskers Bcquest Ansxter Incorrect; Try Agaln; attempts remaining Part Wnat the oleciric ilux tnrough the rectangle (140i 200j) N/C? N.m? /C Submil Previous Answers Correct Here prTD that when the field lines are Carallg Funacp electric ilux tnrough this surface_ equals _



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DRAWING CONCLUSIONShe vertices of a rectangle
are $\mathrm{Q}(2,-3), \mathrm{R}(2,4), \mathrm{S}(5,4),$ and $\mathrm{T}(5,-3) .$
a. Translate rectangle QRST 3 units left and 3 units down to produce rectangle $\mathrm{Q}^{\prime} \mathrm{R}^{\prime} \mathrm{S}^{\prime} \mathrm{T}^{\prime} .$ Find thearea of rectangle QRST and the area of rectangle Q'R'S'T'.
b. Compare the areas. Make a conjecture about the areas of a preimage and its image after a translation.

For this problem. On the topic of electric potential, we have four identical point charges, each with magnitude positive 10 micro columns located on the corners of a rectangle. As we can see in the figure. The rectangle has dimensions of 60 centimeters by 15 centimeters. We want to find the electric potential energy of the charge at the lower left corner due to the other three charges. So the electric potential energy U. E. Is equal to you for V. One plus Q. For the two. Let's cue for the three due to the other the charges. And this is equal to the charge at the lower left corner. Q four into one over four pi. Absolutely not into Q. One of our one les que tu over our two plus Q three divided by its distance to the lower left corner are three. And so since all of these values are known, we can put them in, this is the charge Q four, which is 10 times 10 to the minus six columns and build suppress the units here and that's squared into 1/4 pipes or not, which is the electric constant 8.99 times 10 to the nine newton meters squared. McCullum squared all of this into one over zero 0.6 m plus one over 0.1 five m plus one over the square root of zero point six m squared plus 0.1 five m squared. And so calculating, we get the electric potential energy at the lower left corner to be eight 0.95 jewels.

If this problem you want solved for the potential electric energy of the system. So that is a combination of charge for Wednesday. One plus charge for Q four times B two plus charge four times B three. So simplified here that gives you one over for pie Absalon zero times Q. One over our one plus que two over our two plus Q three over our three information's pulled from the diagram. So if you want Arar one plus Q. Two o r. Two plus Q three over our three. So we simplify that and plug in the values so we get potential energy. Electric energy equals 10 times tended make of six Coombs squared times a point 99 times 10 to the ninth Duden We're squared who squared, then multiplied by for the position here. One over 0.6 meters plus one over 0.150 meters, plus one over this square room 0.6 meters squared plus 0.15 meters quantity squared, so simplify multiply every through yeah, electric energy potential of eight point 95 kills

Hi. In the given problem, first of all, there are coordinate taxes which are given us here. This is why this is that. And this is X axis. There is a cube whose ages are along these according it access in that cube. This is the bottom most. This is the back walls. Place the backward face off this cube. Then the front most face is this. This one is the front most face, then the top most and the bottom most face. So these are the six faces which are named As for the left site. This is S one for the right side. This is s three for the top most face. This is as to for the bottom most face This is S four for this front face. This is s five. And for the back face. This is s six. The electric fields in better form has been given S E is equal to minus 5100 Newton Park Coolum into Meter X into icap Yes, 3.0 Newton Park Coolum meter into the set cake. So it is clear that these components off the electric field are along X axis and alone. Negative X axis. So this is represented like this. This is the component off electric field along negative X axis and another component off electric field is along positive that axis like this. So these are the two components off electric fields and their values are given us for e X. This is minus 5.0 X newton for column into meter. So it is variable with the distance along X axis and for is that this has given us 3.0 Newton into Zet, Newton for Coolum into meter again. This is also variable Peter distance alongside access. Now we will find the electric flux dream through all these surfaces in the first part of the problem. So for as one first will follow as both the components off electric field are parallel to the surface is s one and s three. So no flux will be linked through it. So as e x and E vai is at both are parallel toe s one, so we can say flux linked through this F five flux linked through this s one means five as one is zero. Now it will look for as to here. This is as to the top most part, and the component E Zet is going out off it, so there will be flux. Lean through it five. As to which is given by the product off electric field means 3.0 Zet Newton per column into meter, multiplied by the distance in place of desert. This little distance off this surface from the origin or and that distance will be equal to the length of this cube, which is 0.3 m. So here, in place off this that we will put the value off this distance and which is zero point three meter so Newton or column into meter. This meter will be canceled. So this is the net value of electric field at this surface as to multiplied by the area, an area off each face off. This cube is the square off 0.3 means 0.9 m square, so the multiplied by 0.9 m square, so this electric flux linked through the surface as to comes out Toby 0.81 Newton need to re square for Fulham now for a street again. The flux linked through this will come out to be zero, as we have discussed earlier. Uh, electric field components. Both the components are parallel toe this surface, also as three. Also, So no flux will be linked through it now, for as for this surface is the bottom. Most surface and electric field line means this Zet component off electric field is entering into it, so it will be taken as a negative. But then we look for that distance. Zed, off this surface from the origin, this is zero. So as zero is zero, so the electric field here will be zero. So the flux five s four will come out to be zero, then for as five, this surface s five is the front most surface. This one whose distance from the origin again becomes equals to 0.3, and the electric field is entering into it, so it will be taken as negative. So here it it will become five as five is equal toe minus fight 0.0 into the distance. 0.3 into the area, 0.9 So finally, this electric flux passing through the surface s five comes out Toby minus five point minus 0.135 Newton Meters square for column at last For surface as six. Here, this is the S six office, the bar, the backward surface And as the distance off, this s six is also zero along the X axis, so x zero for this surface so e x will come out to be zero. Hence, we can say five s six will be zero. So finally, the total flux linked through the skill will become five as one plus five as to plus five as three plus five as four plus five as five and five as six. So when we put all these values, we get a net value as minus 5.4 in tow. 10 dish part minus two Newton meters square per column. And this becomes the answer for the first part off the problem. No. For the second part of the problem, we have to find the total charge and closed within this cube, for which we use the concept that the total flux linked through our close surfaces, given by one upon excellent, no times that charge and closed. So from here, discharge comes out to be absolute. No times that total flux. So it is given by 8.854 in tow. 10 Dish par minus 12 Colon square for Newton into meters were multiplied by the flux, which waas minus 5.4 into 10. Dish bar, minus two Newton meters square per Poland. So finally, discharge comes out. Toby minus 4.78 into 10. Dish par minus 13 Gulen. And this becomes the answer for the second part off the problem. Thank you.

For number 19 were to find the area of this rhombus. Um, since a rhombus has signals make perpendicular with the formula that the areas half the product of the lengths of the diagonals works so half the product of the lengths. So 1/2 the product of the lengths the length of this is 2020 40. The length of this is 2121 42. So you get that? My area is 840.


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