Concept explainers
(a)
The electric potential due to the two charges at the origin.
(a)
Answer to Problem 27PQ
The electric potential due to the two charges at the origin is
Explanation of Solution
The total electric potential at origin is the sum of the potential from individual charges.
Write the expression for total electric potential.
Here,
Conclusion:
Substitute,
Therefore, the electric potential due to the two charges at the origin is
(b)
The electric potential due to the two charges at
(b)
Answer to Problem 27PQ
The electric potential due to the two charges at
Explanation of Solution
The first
Write the expression for the distance between two points if the coordinates are given.
For
Use equation (I) to find the potential due to two charges at
Conclusion:
Substitute,
Substitute,
Substitute,
Therefore, the electric potential due to the two charges at
Want to see more full solutions like this?
Chapter 26 Solutions
Physics for Scientists and Engineers: Foundations and Connections
- A 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_forwardThe three charged particles in Figure P20.11 are at the vertices of an isosceles triangle (where d = 2.00 cm). Taking q = 7.00 C, calculate the electric potential at point A, the midpoint of the base. Figure P20.11arrow_forward
- The three charged particles in Figure P25.22 are at the vertices of an isosceles triangle (where d = 2.00 cm). Taking q = 7.00 C, calculate the electric potential at point A, the midpoint of the base.arrow_forwardAt a certain distance from a charged particle, the magnitude of the electric field is 500 V/m and the electric potential is 3.00 kV. (a) What is the distance to the particle? (b) What is the magnitude of the charge?arrow_forwardA small spherical pith ball of radius 0.50 cm is painted with a silver paint and then -10 C of charge is placed on it. The charged pith ball is put at the center of a gold spherical shell of inner radius 2.0 cm and outer radius 2.2 cm. (a) Find the electric potential of the gold shell with respect to zero potential at infinity, (b) How much charge should you put on the gold shell if you want to make its potential 100 V?arrow_forward
- The two charges in Figure P16.12 are separated by d = 2.00 cm. Find the electric potential at (a) point A and (b) point B, which is hallway between the charges. Figure P16.12arrow_forwardAt a certain distance from a charged particle, the magnitude of the electric field is 500 V/m and the electric potential is 3.00 kV. (a) What is the distance to the particle? (b) What is the magnitude of the charge?arrow_forwardAn electron moving parallel to the x axis has an initial speed of 3.70 106 m/s at the origin. Its speed is reduced to 1.40 105 m/s at the point x = 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. (b) Which point is at the higher potential?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 P26.80 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_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
- College PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning