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- 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_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) 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 = 9.00 cm and uniform linear charge density λ = +7.57 pC/m. Take V = 0 at infinity. What is V at point P at distance d = 5.20 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 7.57 pC/m. With V = 0 at infinity, what is V at P? (a) Number i (b) Number i ·+· -L/2 (a) Units Units L/2 +‡ ‡ ‡+3= L/2 .Р (b) L/2arrow_forwardAsaparrow_forwardIn the figure a small, nonconducting ball of mass m = 1.1 mg and charge q = 1.9 × 108 C (distributed uniformly through its volume) hangs from an insulating thread that makes an angle 0 = 23° 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 Unitsarrow_forward
- (a) Figure (a) shows a nonconducting rod of length L = 5.80 cm and uniform linear charge density λ = +4.87 pC/m. Take V = 0 at infinity. What is Vat point P at distance d= 7.50 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 4.87 pC/m. With V=0 at infinity, what is Vat P? L/2 (a) -L/2 L/21/2- (b)arrow_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_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
- Given water's mass of 18g/mole and the value of the fundamental charge (charge magnitude of the electron and proton), use the largest charge density from the article [+0.1988 Q/M (nC/g)] to determine what fraction of water molecules became ionized (charged) due to triboelectric effects when it flows through the material that causes the largest charge transfer. Give your answer in e/molecule, or electrons transferred per molecule of water. For instance, a value of 0.2 means only one in five molecules of water loses an electron, or that 0.2=20% of water molecules become charged. By the way, the answer is nowhere near that large!arrow_forwardThe Earth has an inwardly directed electric field that varies slightly depending on location and altitude. Make the simplifying assumption that this field is a constant 135 N/C and directed toward Earth’s center. What is the net charge, in Coulombs, on the Earth’s surface?arrow_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
- 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