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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 + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance.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 + 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:arrow_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 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: 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 + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance.arrow_forwardA circular ring of radius 23.0 cm has a continuous charge distribution of -1.4 C/m. How many excess electrons are on the ring? Write your result as multiplicative of 1018. Your result must contain one figure after the decimal point. Maximum of 3% of error is accepted in your answer. One electron charge is -1.6x10 19 C.arrow_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 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:arrow_forwardA line of charge of length L = 42 cm with charge Q = 430.0 nAies along the positive Y axis whose one end is at the origin O. A point charge q = 310.0 µClies on point P = (41, 21.0)Here the coordinates are given in centi-meters. a) Find the Electric field at Pdue to the rod. x component of E Give your answer to at least three significance digits. N/C y component of E Give your answer to at least three significance digits. N/C b)Find the charge density of the rod? why is it in C/mbut not in C/m?? Charge density Give your answer to at least three significance digits. C/marrow_forwardE(r) = Bzi - ax'y'j+ (6@yxy + Bóz³) k %3D where a, B, y and & are constants and r = xi+yj+zk. Find the corresponding charge density p(r).arrow_forward
- Consider the electric field of a point charge.At point A the electric field strength is 9 V/m, at point B the electric field strength is 36 V/m. We measure the strength of the electric eld at the midpoint of the interval AB. What can be the measured value?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 = aer/ao + B/r + bo %3D 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 = | S"Edr= - [ *Edr Calculating the antiderivative or indefinite integral, Vab = (-aage-r/ao + 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/( (e-rb/ao - eralao) + B In( ) + bo ( ))arrow_forwardA point charge q = 2C is situated at a long distance r = 15 m on axis from one end of a thin no conducting rod of length L = 2m (r > L) having a charge Q = 7C (Uniformly distributed along its length). The magnitude of electric fied at the point charge q will feel is: Magnitude of the Electric field q will feel Give your answer up to at least three significance digits. V/marrow_forward
- College PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax CollegePhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning