A rod 12.0 cm long is uniformly charged and has a total charge of -23.0 pC. Determine the magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center. -2.12 x magnitude What is the general expression for the electric field along the axis of a uniform rod? N/C direction toward the rod ov Need Help? Read It Watch It
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Q: A rod 10.0 cm long is uniformly charged and has a total charge of -23.0 pc. Determine the magnitude…
A: Given data- Length of the rod l=10.0 cm =10.0×10-2 m Charge of the rod q=-23.0 μC =-23.0×10-6…
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- A point charge gives rise to an electric field with magnitude 4 N/C at a distance of 8 m. If the distance is increased to 16 m, then what will be the new magnitude of the electric field? N/CA proton is projected in the positive x direction into a region of uniform electric field E = (-5.30 x 105) i N/C at t = 0. The proton travels 6.40 cm as it comes to rest. (a) Determine the acceleration of the proton. 0.9638e7 X magnitude How..do.you find the acceleration of an object if you know the net force that acts on it? m/s² direction --Select- (b) Determine the initial speed of the proton. 0.351e7 X magnitude The electric field is constant, so the force is constant, which means the acceleration will be constant. m/s direction -Select-- (c) Determine the time interval over which the proton comes to rest. 0.3640 You appear to have calculated the time correctly using your incorrect results from parts (a) and (b). s6| 17 aölJI - 2 uacl X 10 uluilg (-) X Content Google dai X 9 https://blackboard.uob.edu.bh/ultra/courses/ 24065 1/cl/outline Not syncing Remaining Time: 29 minutes, 53 seconds. * Question Completion Status: Consider a uniformly charged ring of radius R-0.2 m and linear charge density 2-3 µC/m as shown in the figure. Find the electric potential (in V) at (p) a distance 0.7 m from the center. 0 0.7 m 3.1x 10* 6.2 x 104 O9.3x 104 7.8x 104 O 4.7x10 7:05 PN 4/3/202 arch
- A point charge gives rise to an electric field with magnitude 4 N/C at a distance of 9 m. If the distance is increased to 45 m, then what will be the new magnitude of the electric field? N/CReview | Const Imagine a right angle triangle with two equal sides of 10 cm. Two equal charges of +10μC are placed in the vertices corresponding to the 45° angles. A negative, -20μC charge is placed at the 90° vertex. Find the electric field in the middle of the hypotenuse. You need to provide the magnitude of the field and the direction. For the direction use the angle between the electric field and the hypotenuse (the one less than 90°). Part A - What is the magnitude of the field at the point in the midle of the hypothenuse? μA Units Request Answer Part B - What is the angle between the total field and the hypothenuse? (If the angle is not 90 degress, give the angle which is less than 90 degrees) < Return to Assignment Provide Feedback Value SubmitA point charge Q1 = -10.0 nC is at the origin, and a second point charge Q2 = +12.0 nC is on the x axis at x = 0.500 m. Consider the electric fields E, and E, at x = -1.00 m, generated by Q1 and Q2, respectively. Which of the following is true? A. E1 points to the right, E2 points to the left. В. Ei points to the left, E2 points to the right. С. E, points to the left, E, points to the left. D. E1 points to the right, E2 points to the right.
- Charge q₁1.50 nC is at x₁ = 0 and charge 423.00 nC is at x₂ - 2.00 m. At what point between the two charges is the electric field equal to zero? (Enter the x coordinate in m.) HINT x m MacBook Air IQuestion in the picture= + 4.0 nc located at the origin of a consider an electric dipole consisting of a point charge q1 coordinate system and a second charge q2 = - 4.0 nc located at (0.10 m, 0 m). 0.15 m 0.05 m 0.05 m q1 q2 a) find the electric field at points (0.05 ,0 m) and (0.05 m, 0.15 m) due to this dipole. b) what is the electric force on a 2.0 nc charge placed at these points. c) use the far-field approximation to find the electric field e at a point 1.5 m away from the dipole along the perpendicular bisector at coordinates (0.05 m, 1.5 m)?