Module 3 Problem 1

(11%) Problem 1: Two points, point 1 and point 2, are located inside a region with an electric field pointing to the left, as shown in the figure.\f& 50% Part (a) If a proton is moved from point 1 to point 2, how will the potential energy of the charge-field system change? How will the potential change? O Increases / Increases Increases / Decreases O Decreases / Increases Grade Summary Deductions 096 Decreases / Decreases Potential 10096 Submissions Attempts remaining: 3 (339% per attempt) detailed view Submit Hint Feedback I give up! Hints: 09% deduction per hint. Hints remaining: 3 Feedback: 09% deduction per feedback.50% Part (b) If, instead, an electron is moved from point 1 to point 2, how will the potential energy of the charge-field system change? How will the potential change? Decreases / Increases O Increases / Decreases O Increases / Increases Grade Summary Deductions 046 ODecreases / Decreases Potential 10046 Submissions Attempts remaining: 3 (339% per attempt) detailed view Submit Hint Feedback I give up! Hints: 096 deduction per hint. Hints remaining: 3 Feedback: 09% deduction per feedback.(11%) Problem 2: A point mass of charge q > 0 moves at 25' to a uniform electric field E for a total distance d.33% Part (a) Select a diagram that follows from the problem description. Grade Summary O Deductions 096 Potential 10 09% y,=0 Submissions Attempts remaining: 3 (3396 per attempt) detailed view E O y,= 0 E Submit Hint Feedback I give up! Hints: 0 for a 096 deduction. Hints remaining: _0 Feedback: 09% deduction per feedback.& 33% Part (b) The charge q travels from point yy to point yy, where the total traveled distance is d. Find the potential difference that the charge traverses. When you enter your answer you can assume that yy = 0. Grade Summary Deductions Potential 10046 cos( 0) COB(@) COs(e) 7 9 HOME Submissions Attempts remaining: 10 sin(o) sin(@) (A)UIS 4 5 6 (506 per attempt) 1 2 3 detailed view E END h m BACKSPACE CLEAR Submit Hint Feedback I give up! Hints: 096 deduction per hint. Hints remaining 1 Feedback: 09% deduction per feedback.& 33% Part (c) If the charge q = 9.64 uC, distance d= 150.5 m and the field magnitude E = 36.6 V/m, what is the work needed to keep the charge on its angled trajectory? Answer in joules. W = Grade Summary Deductions 046 Potential 1009% sin( tan() 7 9 HOME Submissions Attempts remaining: 10 cotan() asin() aCOB() F 4 (596 per attempt) atan( acotan() sinh() 1 2 3 detailed view coshO tanh() cotanh( END O Degrees O Radians BACKSPACE CLEAR Submit Hint Feedback I give up! Hints: 096 deduction per hint. Hints remaining: 3 Feedback: 09% deduction per feedback.(119%) Problem 3: Consider the parallel-plate capacitor shown in the figure. The plate separation is 3.5 mm and the the electric field inside is 25 N/C. An electron is positioned halfway between the plates and is given some initial velocity, vi. Randomized Variables d = 3.5 mm E = 25 N/C\f& 50% Part (a) What speed, in meters per second, must the electron have in order to make it to the negatively charged plate? Grade Summary Via Deductions 046 Potential 10096 tan() 7 HOME Submissions Attempts remaining: 10 cotan( asin() acoso E 4 (596 per attempt) atan() acotan() zinho 1 2 3 detailed view coshO tanh() cotanh( 0 END ODegrees O Radians BACKSPACE INII. CLEAR Submit Hint Feedback I give up! Hints: 096 deduction per hint. Hints remaining 1 Feedback: 09% deduction per feedback.50% Part (b) If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What will its speed be, in meters per second, when it reaches the positive plate in this case? VEb Grade Summary Deductions 096 Potential 10046 sin() tan() HOME Submissions cotan( asin() acos( E Attempts remaining: 10 (50% per attempt) atan( acotan() zinho EC 1 2 3 detailed view coshO tanh() cotanh() + 0 END O Degrees O Radians BACKSPACE CLEAR Submit Hint Feedback I give up! Hints: 096 deduction per hint. Hints remaining: 1 Feedback: 09% deduction per feedback.(11%) Problem 4: The electric field in a certain region is given by the function E = Ak cos (kx) cos (by) i - Absin (kr) sin (by) j where A = 5.86 N - m/C, k =0.802 m , and b = 1.39 m . The points in the figure use the values T1 = 1.75 m and y = 2.61 m.\f& 17% Part (a) What is the change in electric potential, in volts, from point (0, 0) to point (x1, 0)? Grade Summary V(x1, 0) - 10, 0) = Deductions 096 Potential 1009% sin() COs( tan() 7 HOME Submissions Attempts remaining: 10 cotan( ) asin() acos() F A (596 per attempt) atan() sinh() 1 2 detailed view acotan() 3 cosh() tanh() cotanh() + END O Degrees O Radians HACKSPACE CLEAR Submit Hint Feedback I give up! Hints: 09% deduction per hint. Hints remaining: 3 Feedback: 096 deduction per feedback.17% Part (b) What is the change in potential, in volts, from point (x1: 0) to point (x1, y)? Grade Summary Wxy) - V(x1 0) = V Deductions Potential 1009% sin() cos() tan( 1 7 9 HOME Submissions Attempts remaining 10 cotan( ) acos() F 4 (596 per attempt) atan( ) acotan() sinh() 1 13 detailed view cosh( tanh() cotanh() T END O Degrees O Radians NO BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 09% deduction per hint. Hints remaining: 1 Feedback: 09% deduction per feedback.17% Part (c) What is the change in potential, in volts, from point (0, 0) to point (x1. )1), along the path that passes through (x1; 0)? Grade Summary Vry) - 10, 0) = V Deductions 096 Potential 1009% sin() cost tan() 7 HOME Submissions Attempts remaining 10 cotan() asin() acos() F 5 6 (596 per attempt) atan() acotan() sinh() 1 2 3 detailed view cosh( tanh() cotanh() END O Degree: O Radians NO HACKSPACE CLEAR Submit Hint Feedback I give up! Hints: 09% deduction per hint. Hints remaining: 1 Feedback: 096 deduction per feedback.& 17% Part (d) What is the change in electric potential, in volts, from point (0, 0) to point (0, y)? Grade Summary W(0, y1) - 10, 0) = V Deductions Potential 1009% sin() COs() tan() 7 HOME Submissions Attempts remaining 10 cotan() asinO acos() F A 5 (596 per attempt) 1 2 detailed view atan( ) acotan() sinh( 3 cosh() tanh() cotanh() END O Degrees O Radians NO BACKSPACE CLEAR Submit Hint Feedback I give up! Hints: 09% deduction per hint. Hints remaining: 1 Feedback: 096 deduction per feedback.17% Part (e) What is the change in electric potential, in volts, from point (0, y) to point (x1: y1)? Grade Summary Deductions 096 Potential 1009% sin() cosQ tan( 7 9 HOME Submissions Attempts remaining 10 cotan() asin( acoso E 4 (596 per attempt) atan() acotan() smh() 1 2 3 detailed view cosh( ) tanh() cotanh( ) T END O Degrees O Radians VO BACKSPACE CLEAR Submit Hint Feedback I give up! Hints: 09% deduction per hint. Hints remaining: 1 Feedback: 096 deduction per feedback.&17% Part (f) Find the value of the change in potential, in volts, from point (0, 0) to point (x1: ))- Grade Summary Vx1: y) - 10, 0) = Deductions 096 Potential 1009% sin( COSO tan() 7 HOME Submissions Attempts remaining 10 cotan() asin() acoso 5 (596 per attempt) atan() acotan() sinh( 1 2 3 detailed view cosh() tanh() cotanh() + END O Degrees O Radians VO BACKSPACE CLEAR Submit Hint Feedback I give up! Hints: 09% deduction per hint. Hints remaining: 1 Feedback: 096 deduction per feedback.(11%) Problem 5: Charges move in an electric field due to the Coulomb force.& Some unknown charge moves in a direction perpendicular to the direction of the electric field lines. We can conclude that: Grade Summary Both its electric potential and electric potential energy stay constant. Deductions 046 It moves from a low potential to a high potential, and its electric energy decreases. Potential 10046 It moves from a high potential to a low potential, and its electric energy increases. Submissions It moves from a high potential to a low potential, and its electric energy decreases. Attempts remaining: 4 It moves from a low potential to a high potential, and its electric energy increases. (250% per attempt) detailed view Submit Hint Feedback I give up! Hints: 096 deduction per hint. Hints remaining: Feedback: 09% deduction per feedback.(119%) Problem 6: Consider a proton in a uniform electric field directed left to right, as shown in the figure. For both paths the initial speed of the proton is the same, but the direction of the initial velocity is different.\f50% Part (a) Compare the change in electric potential energy along path A to the change in electric potential energy along path B. Grade Summary OAUA = 4UB Deductions 006 OAUA > AUB Potential 10096 OThere is not enough information given - we need either the initial speed or the size of the electric field. Submissions OAUA

Answer rating: 100% (QA)

The Answer is in the image, click to view ... View the full answer