A rod of length L (Fig. P24.25) lies along the x axis with its left end at the origin. It has a nonuniform charge density 1 = ax, where a is a positive constant. (a) What are the units of a? (b) Calculate the electric potential at A. B. b d A
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- A uniform external electric field points to the right. An electron ( me = 9.11E[-31]kg; qe = –1.602 E[–19]C) is place in the field and fired to the right at v¡ 100, 000m/s. a. Draw a picture of the situation b. The electron will eventually come to a stop. What is the change in potential between the electron's starting point and its stopping point? Remember to submit a pdf of your work on this problem to the Moodle assignment listed in the instructions. Also, answer True for this question.a. Calculate the potential difference Vo - Va due to the infinite rod of uniform charge density X, as shown below. Perform this calculation by directly integrating the electric field over a path between a and b. I need to show all works step by step like physics and math4 charges, each 2 nC worth of charge, are arranged on the 4 corners of a square. The side length of the square is 1 meter. What is the electric potential in the center?
- n a single plane lies a stationary, positive point charge, qs, with two corresponding equipotential curves in the form of circles that both have a common center at the location of qs. The radial distance between the two equipotential circles is 4.0 cm. (See the diagram below.) What is the value for qs? Assume that k = 9.0 x 109 Nm2/C2. a. 3.3 x 10-10 C b. 4.4 x 10-10 C c. 2.0 x 10-12 C d. 4.4 x 10-9 C e. 3.3 x 10-11 CA parallel-plate capacitor has an area of 2.00 cm2, and the plates are separated by 2.00 mm with air between them. The capacitor stores a charge of 500 pC. (a) What is the potential difference across the plates of the capacitor? V(b) What is the magnitude of the uniform electric field in the region between the plates? N/CPlease Don't use Al solution
- The three charges in the figure below are at the vertices of an isosceles triangle. Let q = 6.50 nC and calculate the electric potential at the midpoint of the base. (Let d, = 1.50 cm and d, = 7.00 cm.) d2 d1 -20436.21 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. kVE7BP1