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[20%) Problem 2: Apoin't charge of5.8 [1C is placed at the origin (x1 = D) ofa coordinate system, and another charge ofI.9 pC is placed placed on the x axis at X; = 0.21 m. E & 50% Part (a) Where on the xaxis can a third charge be placed in meters so that the net force on it is zero? E> & 50% Part (I1) What ifboth charges are positive; that is, what ifthe second charge is 1.9 pC? Grade Summary Deductions D % n ______ : _. u M n x. I'3ZI (20%) Problem 3: In the figure, the point charges are located at the corners of an equilateral triangle 22 cm on a side. O qa O qc Otheexpertta.com & 25% Part (a) Find the magnitude of the electric field in N/C at the location of qa: given that q1 = 10.4 uC and q. =-5.9 uC . Eal = sin() cos() tan() 7 8 9 HOME cotan( asin( acos() 5 6 atan( acotan( ) sinh() 1 2 3 cosh( tanh() cotanh() + 0 END O Degrees O Radians NO BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 4 Feedback: 1% deduction per feedback A 25% Part (b) Find the direction of the electric field at qa in degrees above the negative x-axis with origin at qa- A 25% Part (c) What is the magnitude of the force in N on qa: given that qa = 1.9 nC? 4 25% Part (d) What is the direction of the force on qa in degrees above the negative x-axis with origin at qa?(20%) Problem 4: An electron is orbiting a nucleus which has a charge of 16e, under the action of the Coulomb force at a radius of 1.66 x 10-10 m. A Calculate the angular velocity of the electron, in radians per second. (0 = F t: sin( cos() tan() 7 HOME cotan() asin() acos() E TA atan( acotan() sinhC 3 coshQ tanh() cotanh() + 0 END O Degrees O Radians VO BACKSPACE DEL CLEAR20%) Problem 5: An electron has an initial velocity of 4.8 x 10 m/s in a uniform 1.9 x 10 N/C electric field. The field accelerates the electron in he direction opposite to its initial velocity. 25% Part (a) What is the direction of the electric field? Grade = 34% Correct Answer Student Final Submission Feedback The field is in the direction of the electron's initial The field is in the direction of the electron's initial Correct! velocity. velocity Grade Summary Deduction for Final Submission 0% Deductions for Incorrect Submissions, Hints and Feedback [?] 66% Student Grade = 100 - 0 - 66 = 34% Submission History All Date times are displayed in Eastern Standard Time Red submission date times indicate late work. Date Time Answer Hints Feedback Feb 27, 2022 12:49 PM The field is in the opposite direction of the electron's initial velocity. Note: Feedback not accessed. 2 Feb 27, 2022 12:49 PM The field is in the direction to the right of the electron's initial velocity. Note: Feedback not accessed. 3 Feb 27, 2022 12:50 PM The field is in the direction of the electron's initial velocity. * 25% Part (b) How far does the electron travel before coming to rest in m? x = 5.57 * 10-3 x = 5.57E-5 X Attempts Remain * 25% Part (c) How long does it take the electron to come to rest in s? *= 24.2 X Attempts Remain * 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 elocity? vf = 4.1 * 106 vf = 4100000 X Attempts Remain