Concept explainers
Two very large horizontal sheets are 4.25 cm apart and carry equal but opposite uniform surface charge densities of magnitude a. You want to use these sheets to hold stationary in the region between them an oil droplet of mass 486 μg that carries an excess of five electrons. Assuming that the drop is in vacuum, (a) which way should the electric field between the plates point, and (b) what should σ be?
Learn your wayIncludes step-by-step video
Chapter 21 Solutions
University Physics with Modern Physics Plus Mastering Physics with eText -- Access Card Package (14th Edition)
Additional Science Textbook Solutions
Conceptual Physics (12th Edition)
Conceptual Physical Science (6th Edition)
Physics: Principles with Applications
College Physics: A Strategic Approach (4th Edition)
Essential University Physics: Volume 1 (3rd Edition)
- Two solid spheres, both of radius 5 cm, carry identical total charges of 2 C. Sphere A is a good conductor. Sphere B is an insulator, and its charge is distributed uniformly throughout its volume. (i) How do the magnitudes of the electric fields they separately create at a radial distance of 6 cm compare? (a) EA EB = 0 (b) EA EB 0 (c) EA = EB 0 (d) 0 EA EB (e) 0 = EA EB (ii) How do the magnitudes of the electric fields they separately create at radius 4 cm compare? Choose from the same possibilities as in part (i).arrow_forwardaA plastic rod of length = 24.0 cm is uniformly charged with a total charge of +12.0 C. The rod is formed into a semicircle with its center at the origin of the xy plane (Fig. P24.34). What are the magnitude and direction of the electric field at the origin? Figure P24.34arrow_forwardIn the figure a small, nonconducting ball of mass m = 1.3 mg and charge q = 1.7 × 10° C (distributed uniformly through its volume) hangs from an insulating thread that makes an angle e= 41° with a vertical, uniformly charged nonconducting sheet (shown in cross section). Considering the gravitational force on the ball and assuming the sheet extends far vertically and into and out of the page, calculate the surface charge density o of the sheet. m, 4 Number i Units > +++ +++F++arrow_forward
- An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity o1 = 0.62 µC/m2. Another infinite sheet of charge with uniform charge density o7 = -0.29 µC/m? is located at x = c = 33 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 14.5 cm and x 3 18.5 сm). a/2 a/2| a/2 1) What is Ex(P), the x-component of the electric field at point P, located at (x,y) = (7.25 cm, 0)? N/C Submit 2) What is oa, the charge density on the surface of the conducting slab at x = 14.5 cm? µC/m? Submit 3) What is V(R) - V(P), the potentital difference between point P and point R, located at (x,y) = (7.25 cm, -18.5 cm)? Submit 4) What is V(S) - V(P), the potentital difference between point P and point S, located at (x,y) = (25.75 cm, -18.5 cm)? Submit + 5) What is Ex(T), the x-component of the electric field at point T, located at (x,y) (40.25 сm, -18.5 ст)? N/C Submitarrow_forwardTwo infinite sheets plates with the same uniform charge density of 8.0 µC/m2 are vertically placed and parallel to each other. a distance d apart along the yz-plane. One plane passes through x = 3 cm while the other plate passes through x = -3 cm. The magnitude of the net electric field at the point (x = 5, y = 0, z = 0) is nearly Question 3 options: None of these 9.0 x 105 N/C 0 4.5 x 105 N/Carrow_forwardIn the figure a small, nonconducting ball of mass m = 1.1 mg and charge q = 2.4 × 10-8 C (distributed uniformly through its volume) hangs from an insulating thread that makes an angle θ = 38° with a vertical, uniformly charged nonconducting sheet (shown in cross section). Considering the gravitational force on the ball and assuming the sheet extends far vertically and into and out of the page, calculate the surface charge density σ of the sheet.arrow_forward
- (a) Figure (a) shows a nonconducting rod of length L-5.10 cm and uniform linear charge density = +8.35 pC/m. Take V = 0 at infinity. What is V at point P at distance d = 7.60 cm along the rod's perpendicular bisector? (b) Figure (b) shows an identical rod except that one half is now negatively charged. Both halves have a linear charge density of magnitude 8.35 pC/m. With V-0 at infinity, what is V at P? P L/2 L/2- (a) L/21/2 (b) (a) Number i Units (b) Number i Unitsarrow_forward(a) Figure (a) shows a nonconducting rod of length L-5.20 cm and uniform linear charge density A= +5.99 pC/m. Take V = 0 at infinity. What is Vat point P at distance d = 8.20 cm along the rod's perpendicular bisector? (b) Figure (b) shows an identical rod except that one half is now negatively charged. Both halves have a linear charge density of magnitude 5.99 pC/m. With V 0 at infinity, what is Vat P? L/2 L/2 –L/2 L/2- (a) (b) (a) Number Units V (b) Number Units Varrow_forward(a) Figure (a) shows a nonconducting rod of length L = 5.40 cm and uniform linear charge density λ = +4.41 pC/m. Take V = 0 at infinity. What is V at point P at distance d = 9.30 cm along the rod's perpendicular bisector? (b) Figure (b) shows an identical rod except that one half is now negatively charged. Both halves have a linear charge density of magnitude 4.41 pC/m. With V= 0 at infinity, what is V at P? (a) Number i (b) Number i P ‡ ‡ ‡ ‡ + + + +‡‡ ‡ ‡‡ L/2 L/2 Units Units [+ + + ++++G ·L/2 L/2-arrow_forward
- (a) What total (excess) charge q must the disk have for the electric field on the surface of the disk at its center to have magnitude 3.0* 10^6 N/C, the E value at which air breaks down electrically, producing sparks? Take the disk radius as 2.5 cm. (b) Suppose each surface atom has an effective cross-sectional area of 0.015 nm2. How many atoms are needed to make up the disk surface? (c) The charge calculated in (a) results from some of the surface atoms having one excess electron. What fraction of these atoms must be so charged?arrow_forwardAn infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.41 μC/m2. Another infinite sheet of charge with uniform charge density σ2 = -0.11 μC/m2 is located at x = c = 32 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 14 cm and x = 18 cm). 1) What is Ex(P), the x-component of the electric field at point P, located at (x,y) = (7 cm, 0)? N/C 2) What is σa, the charge density on the surface of the conducting slab at x = 14 cm? μC/m2 3) What is V(R) - V(P), the potentital difference between point P and point R, located at (x,y) = (7 cm, -18 cm)? V 4) What is V(S) - V(P), the potentital difference between point P and point S, located at (x,y) = (25 cm, -18 cm)? V 5) What is Ex(T), the x-component of the electric field at point T, located at (x,y) = (39 cm, -18 cm)? N/Carrow_forward(a) Figure (a) shows a nonconducting rod of length L = 8.00 cm and uniform linear charge density λ = +1.21 pc/m. Take V = 0 at infinity. What is V at point P at distance d = 7.40 cm along the rod's perpendicular bisector? (b) Figure (b) shows an identical rod except that one half is now negatively charged. Both halves have a linear charge density of magnitude 1.21 pc/m. With V = 0 at infinity, what is V at P? 1/2 L/2- L/2 1/2 - (a) (b)arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning