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- A 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.Number 1 part a b and c 1. The element gold, 797 Au, has a density of 19300kg/m?. (a) How many electrons are there in 15g of gold? (Assume that it is no net charge). (b) If this 15g is given a net charge of 25uC, approximately how many electrons were removed? (c) Approximately what fraction of the total number of electrons were removed?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 rp has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = ae-T/ao + B/r + bo where alpha (a), beta (8), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vah between the two conductors. The potential difference is related to the electric field by: Vab = Edr = - Edr Calculating the antiderivative or indefinite integral, Vab = (-aage-r/a0 + 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 ( ))
- 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 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 ( ))The electric field vector is given by E=[0.2, –0.6, 0.3] V/m and Area vector points to A=[–4, 3, -5] m^2. What is the angle between the E-field and Area vector in degrees? Answer with two decimal places. note: convert answer to milliSuppose 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 () )
- 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: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.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 () )
- 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 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 %| 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: Va 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 = Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q/( (e-"b/ao - era/ao) + B In( ) + bo ( ))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 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 ( ))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 ( ))