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DATA Positive charge Q is distributed uniformly around a very thin
Figure P21.94
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University Physics with Modern Physics, Books a la Carte Edition; Modified MasteringPhysics with Pearson eText -- ValuePack Access Card -- for ... eText -- Valuepack Access Card (14th Edition)
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- Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Calculating the antiderivative or indefinite integral , Vab = (-αa0e-r/a0 + β + b0 ) By definition, the capacitance C is related to the charge and potential difference by: C = / Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q / ( (e-rb/a0 - e-ra/a0) + β ln() + b0 () )arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0arrow_forwardA thick insulating spherical shell of inner radius a=2.4R and outer radius b=6.1R has a uniform charge density p. pR What is the magnitude of the electric field at r=5.6 R ? Express your answer using one decimal place in units of €oarrow_forward
- Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rh has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = ae-r/ao + B/r + bo where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Vab = Edr = - Edr Calculating the antiderivative or indefinite integral, Vab = (-aager/ao + B + bo By definition, the capacitance Cis related to the charge and potential difference by: C = Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q/( (e rb/ao - eTalao) + B In( ) + bo ( ))arrow_forwardA 27 nC charge generates an electric field within an isolated chamber. At a point a distance d meters away from the charge, the electric field is measured to be 4 N/C. Determine d in meters. Use k=9 x109 for the electric constant.arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rp has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = ae-r/ao + B/r + bo where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: ['´e Vob = Edr= - Edr Calculating the antiderivative or indefinite integral, Vab = (-aaoe¯r7ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C = Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q / ( (erb/ao - eralao) + B In( ) + bo ( ))arrow_forward
- 21.84 . CALC A semicircle of radius a is in the first and second y. quadrants, with the center of cur- vature at the origin. Positive charge +Q is distributed uniformly around the left half of the semicircle, and negative charge -Q is distributed uniformly around the right half of the semicircle (Fig. P21.84). What are the magnitude and direction of the net electric field at the origin produced by this distribution of charge? +Q x-arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rh has charge -Q. The electric field E at a radial distancer from the central axis is given by the function: E = ae-r/ao + B/r + bo where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Vab = Edr = - Edr Calculating the antiderivative or indefinite integral, Vab = (-aaoe-r/ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C= Q I Vabarrow_forwardIn deep space two spheres each of radius 6.30 m are connected by a 2.28 ✕ 102 m nonconducting cord. If a uniformly distributed charge of 26.5 mC resides on the surface of each sphere, calculate the tension in the cord. Answer Unit is Narrow_forward
- This question checks that you can use the formula of the electric field due to a long, thin wire with charge on it. The field due to an infinitely long, thin wire with linear charge E = 12X Απερ η density is Imagine a long, thin wire with a constant charge per unit length of -2.3×10 C/m. What is the magnitude of the electric field at a point 10 cm from the wire (assuming that the point is much closer to the wire's nearest point than to either of its ends)? Give your answer in units of kN/C. -7arrow_forwardTwo parallel, thin, L×L conducting plates are separated by a distance d. Let L=1.7 m, and d=2.0 mm. A charge of +6.5μC is placed on one plate, and a charge of −6.5μC is placed on the other plate a. What is the magnitude of charge density on the inside surface of each plate, in coulombs per square meter? b. What is the magnitude of the electric field between the plates?arrow_forwardYou are working as an intern for a meteorological laboratory. You are out in the field taking measurements with a device that measures electric fields. You measure the electric field in the air immediately above the Earth's surface to be 139 N/C directed downward. (Assume the radius of the Earth is 6.37 x 106 m.) (a) Determine the surface charge density (in C/m²) on the ground. C/m? (b) Imagine the surface charge density is uniform over the planet. Determine the charge (in C) of the whole surface of the Earth. (e) Determine the Earth's electric potential (in V) due to the charge found in (b). V (d) Determine the difference in potential (in V) between the head and the feet of a person 1.50 m tall. (Ignore any charges in the atmosphere.) Varrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax CollegeCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
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