EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 23, Problem 52P
(a)
To determine
The expression for electric potential.
(b)
To determine
The proof that part (a) can be reduced to
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Chapter 23 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
Ch. 23 - Prob. 1PCh. 23 - Prob. 2PCh. 23 - Prob. 3PCh. 23 - Prob. 4PCh. 23 - Prob. 5PCh. 23 - Prob. 6PCh. 23 - Prob. 7PCh. 23 - Prob. 8PCh. 23 - Prob. 9PCh. 23 - Prob. 10P
Ch. 23 - Prob. 11PCh. 23 - Prob. 12PCh. 23 - Prob. 13PCh. 23 - Prob. 14PCh. 23 - Prob. 15PCh. 23 - Prob. 16PCh. 23 - Prob. 17PCh. 23 - Prob. 18PCh. 23 - Prob. 19PCh. 23 - Prob. 20PCh. 23 - Prob. 21PCh. 23 - Prob. 22PCh. 23 - Prob. 23PCh. 23 - Prob. 24PCh. 23 - Prob. 25PCh. 23 - Prob. 26PCh. 23 - Prob. 27PCh. 23 - Prob. 28PCh. 23 - Prob. 29PCh. 23 - Prob. 30PCh. 23 - Prob. 31PCh. 23 - Prob. 32PCh. 23 - Prob. 33PCh. 23 - Prob. 34PCh. 23 - Prob. 35PCh. 23 - Prob. 36PCh. 23 - Prob. 37PCh. 23 - Prob. 38PCh. 23 - Prob. 39PCh. 23 - Prob. 40PCh. 23 - Prob. 41PCh. 23 - Prob. 42PCh. 23 - Prob. 43PCh. 23 - Prob. 44PCh. 23 - Prob. 45PCh. 23 - Prob. 46PCh. 23 - Prob. 47PCh. 23 - Prob. 48PCh. 23 - Prob. 49PCh. 23 - Prob. 50PCh. 23 - Prob. 51PCh. 23 - Prob. 52PCh. 23 - Prob. 53PCh. 23 - Prob. 54PCh. 23 - Prob. 55PCh. 23 - Prob. 56PCh. 23 - Prob. 57PCh. 23 - Prob. 58PCh. 23 - Prob. 59PCh. 23 - Prob. 60PCh. 23 - Prob. 61PCh. 23 - Prob. 62PCh. 23 - Prob. 63PCh. 23 - Prob. 64PCh. 23 - Prob. 65PCh. 23 - Prob. 66PCh. 23 - Prob. 67PCh. 23 - Prob. 68PCh. 23 - Prob. 69PCh. 23 - Prob. 70PCh. 23 - Prob. 71PCh. 23 - Prob. 72PCh. 23 - Prob. 73PCh. 23 - Prob. 74PCh. 23 - Prob. 75PCh. 23 - Prob. 76PCh. 23 - Prob. 77PCh. 23 - Prob. 78PCh. 23 - Prob. 79PCh. 23 - Prob. 80PCh. 23 - Prob. 81PCh. 23 - Prob. 82PCh. 23 - Prob. 83PCh. 23 - Prob. 84PCh. 23 - Prob. 85PCh. 23 - Prob. 86PCh. 23 - Prob. 87PCh. 23 - Prob. 88PCh. 23 - Prob. 89PCh. 23 - Prob. 90PCh. 23 - Prob. 91PCh. 23 - Prob. 92PCh. 23 - Prob. 93PCh. 23 - Prob. 94P
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- An electric potential exists in a region of space such that V = 8x4 2y2 + 9z3 and V is in units of volts, when x, y, and z are in meters. a. Find an expression for the electric field as a function of position. b. What is the electric field at (2.0 m, 4.5 m, 2.0 m)?arrow_forwardA filament running along the x axis from the origin to x = 80.0 cm carries electric charge with uniform density. At the point P with coordinates (x = 80.0 cm, y = 80.0 cm), this filament creates electric potential 100 V. Now we add another filament along the y axis, running from the origin to y = 80.0 cm, carrying the same amount of charge with the same uniform density. At the same point P, is the electric potential created by the pair of filaments (a) greater than 200 V, (b) 200 V, (c) 100 V, (d) between 0 and 200 V, or (e) 0?arrow_forwardA filament running along the x axis from the origin to x = 80.0 cm carries electric charge with uniform density. At the point P with coordinates (x = 80.0 cm, y = 80.0 cm), this filament creates electric potential 100 V. Now we add another filament along the y axis, running from the origin to y = 80.0 cm. carrying the same amount of charge with the same uniform density. At the same point P, is the electric potential created by the pair of filaments (a) greater than 200 V, (b) 200 V, (c) 100 V, (d) between 0 and 200 V, or (e) 0?arrow_forward
- Two point charges, q1 = 2.0 C and q2 = 2.0 C, are placed on the x axis at x = 1.0 m and x = 1.0 m, respectively (Fig. P26.24). a. What are the electric potentials at the points P (0, 1.0 m) and R (2.0 m, 0)? b. Find the work done in moving a 1.0-C charge from P to R along a straight line joining the two points. c. Is there any path along which the work done in moving the charge from P to R is less than the value from part (b)? Explain.arrow_forwardA source consists of three charged particles located at the vertices of a square (Fig. P26.32), where the square has sides of length 0.243 m. The charges are q1 = 35.0 nC, q2 = 65.0 nC, and q3 = 56.5 nC. Find the electric potential at point A located at the fourth vertex. FIGURE P26.32 Problems 32 and 33.arrow_forwardFigure P26.44 shows a rod of length = 1.00 m aligned with the y axis and oriented so that its lower end is at the origin. The charge density on the rod is given by = a + by, with a = 2.00 C/m2 and b = 1.00 C /m2. What is the electric potential at point P with coordinates (0, 25.0 cm)? A table of integrals will aid you in solving this problem.arrow_forward
- The charge density on a disk of radius R = 12.0 cm is given by = ar, with a = 1.40 C/m3 and r measured radially outward from the origin (Fig. P26.45). What is the electric potential at point A, a distance of 40.0 cm above the disk? Hint: You will need to integrate the nonuniform charge density to find the electric potential. You will find a table of integrals helpful for performing the integration.arrow_forwardFigure P26.80 shows a wire with uniform charge per unit length = 2.25 nC/m comprised of two straight sections of length d = 75.0 cm and a semicircle with radius r = 25.0 cm. What is the electric potential at point P, the center of the semicircular portion of the wire? FIGURE P26.80arrow_forwardAn infinite number of charges with q = 2.0 C are placed along the x axis at x = 1.0 m, x = 2.0 m, x = 4.0 m, x = 8.0 m, and so on, as shown in Figure P26.78. Determine the electric potential at the point x = 0 due to this set of charges. Hint: Use the mathematical formula for a geometric series, 1+r+r2+r3+r4+=11r FIGURE P26.78arrow_forward
- Three particles with equal positive charges q are at the corners of an equilateral triangle of side a as shown in Figure P20.10. (a) At what point, if any, in the plane of the particles is the electric potential zero? (b) What is the electric potential at the position of one of the particles due to the other two particles in the triangle? Figure P20.10arrow_forward(a) Find the electric potential, taking zero at infinity, at the upper right corner (the corner without a charge) of the rectangle in Figure P16.13. (b) Repeat if the 2.00-C charge is replaced with a charge of 2.00 C. Figure P16.13 Problems 13 and 14.arrow_forwardThe electric potential inside a charged spherical conductor of radius R is given by V = keQ/R, and the potential outside is given by V = keQ/R, Using Er = dV/dr, derive the electric field (a) inside and (b) outside this charge distribution.arrow_forward
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