(8%) Problem 1: A point charge of 5.9 uC is placed at the origin (x1 = 0) of a coordinate system, and another charge of -1.5 JC is placed placed on the x-axis at x2 = 0.22 m. 50% Part (a) Where on the x-axis can a third charge be placed in meters so that the net force on it is zero? Grade Summary X3 = Deductions 0% Potential 100% sin() cos tan( 8 9 HOME Submissions Attempts remaining: 100 cotan() asin() acos() A 5 6 0% per attempt atan() acotan() sinh() 2 3 detailed view cosh() tanh( ) cotanh() END O Degrees O Radians VO BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 2% deduction per hint. Hints remaining: 3 Feedback: 2% deduction per feedback. 4 50% Part (b) What if both charges are positive; that is, what if the second charge is 1.5 JC?(8%) Problem 2: Point charges of 0.26 uC and 0.61 uC are placed 0.65 m apart. 50% Part (a) At what point along the line between them is the electric field zero? Give your answer in meters from the 0.26 JC charge. Grade Summary X= Deductions 0% Potential 100% sin() cos() tan 7 8 9 HOME Submissions Attempts remaining: 100 cotan() asin() acos() E A 1 5 6 (0% per attempt) atan() acotan() sinh() 2 3 detailed view cosh( tanh() cotanh() + 0 END O Degrees O Radians VO BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 2% deduction per hint. Hints remaining: 5 Feedback: 2% deduction per feedback. 4 50% Part (b) What is the magnitude of the electric field, in newtons per coulomb, halfway between them?(6%) Problem 3: A simple and common technique for accelerating electrons is shown in the figure, which depicts a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to pass through. Randomized Variables E=2.55 x 104 N/C Otheexpertta.com Calculate the horizontal component of the electron's acceleration if the field strength is 2.55 x 104 N/C. Express your answer in meters per second squared, and assume the electric field is pointing in the negative x-direction as shown in the figure. Grade Summary a = Deductions 0% Potential 100% sin() cos() tan() 7 8 9 HOME Submissions Attempts remaining: 100 cotan() asin() acos() E 4 5 6 (0% per attempt) atan() acotan() sinh() 2 3 detailed view cosh() tanh() cotanh() + END Degrees O Radians VO BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 2% deduction per hint. Hints remaining: 2 Feedback: 2% deduction per feedback.(11%) Problem 4: An electron has an initial velocity of 4.8 x 10 m/s in a uniform 1.85 x 10 N/C electric field. The field accelerates the electron in the direction opposite to its initial velocity. 25% Part (a) What is the direction of the electric field? The field is in the direction of the electron's initial velocity. Correct! 25% Part (b) How far does the electron travel before coming to rest in m? Grade Summary X= Deductions 0% Potential 100% sin() cos() tan() 7 8 9 HOME Submissions Attempts remaining: 100 cotan asin() acos E 4 5 6 (0% per attempt) atan( acotan() sinhC 1 2 3 detailed view cosh() tanh() cotanh() + 0 END Degrees O Radians VO BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 2% deduction per hint. Hints remaining: 3 Feedback: 2% deduction per feedback. A 25% Part (c) How long does it take the electron to come to rest in s? 4 25% Part (d) What is the magnitude of the electron's velocity (in m/s) when it returns to its starting point in the opposite direction of its initial velocity