The wire in the figure to the right has linear charge density 2. What is the electric potential at the center of the semicircle? Give your answer in terms of λ, R, € and appropriate constants. *Hint: you do not need to integrate each segment, you can use results from previous work. Answer: V = 2/(4€)(1+2ln(3)/π)
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- The electric potential in a region of space is V = 300V-m, where and y are in meters. √z²+y² Part A What is the strength of the electric field at (x, y) = (2.9m, 1.3m)? Express your answer in volts per meter. E= Submit Part B 0= V=| ΑΣΦΑ Submit 4 Request Answer IVE ΑΣΦ What is the direction of the electric field at (x, y) = (2.9m, 1.3m)? Give the direction as an angle ccw from the positive x-axis. Express your answer in degrees. ? Request Answer V/m ? ºccw from the positive z-axisLearning Goal: To understand the relationship and differences between electric potential and electric potential energy. In this problem we will learn about the relationships between electric force F, electric field Ē, potential energy U, and electric potential V. To understand these concepts, we will first study a system with which you are already familiar: the uniform gravitational field. Part B Now find the gravitational potential energy U (z) of the object when it is at an arbitrary height z. Take zero potential to be at position z = 0. Keep in mind that the potential energy is a scalar, not a vector. Express U (2) in terms of m, z, and g. U (z) = Submit ΨΕ ΑΣΦ Request Answer ?In the figure below, we see a uniform plane charge distribution in a square of length L = 20 cm.Point p is a distance of 2.0 mm above the plane, near the center of the charge distribution(drawing is not to scale). The distance d indicated in the figure is 5.0 mm.A. Find |∆V| between points P and Q�.B. Which point is at the lower potential? Explain.
- part b plzA 15.0 nC charge is at x = 0cm and a -1.2 nC charge is at x = 7 cm . Part A At what point or points on the x-axis is the electric potential zero? Express your answer using two significant figures. If there is more than one answer, give each answer separated by a comma. Xo = cm Submit Request AnswerPlease refer to the photos
- An alpha particle with a kinetic energy of 7.00 MeV makes a head-on collision with a gold nucleus at rest. Part A What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. The atomic number of gold is 79, and an alpha particle is a helium nucleus consisting of two protons and two neutrons.) Express your answer in meters. d = Submit VE ΑΣΦ Provide Feedback Request Answer ? mA uniform electric field of magnitude 7.8x105 N/C points in the positive z direction. Part A Find the change in electric potential energy of a 8.0-μC charge as it moves from the origin to the point (0, 6.0 m). Express your answer using one significant figure. IVE] ΑΣΦ AU = Submit Part B AU = Find the change in electric potential energy of a 8.0-μC charge as it moves from the origin to the point (6.0 m, 0). Express your answer using two significant figures. IVE] ΑΣΦ Submit Part C Request Answer AU = Submit Request Answer w ? Find the change in electric potential energy of a 8.0-μC charge as it moves from the origin to the point (6.0 m, 6.0 m). Express your answer using two significant figures. [5] ΑΣΦ Request Answer ? w ? J J JNeeds Complete typed solution with 100 % accuracy.
- Part A The work done by an external force to move a -5.60 µC charge from point A to point B is 1.50x10-3 J. If the charge was started from rest and had 4.72x10-4 J of kinetic energy when it reached point B, what must be the potential difference between A and B? Express your answer with the appropriate units. HẢ ? VB – VA = Value UnitsCan you help me to solve this?