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- A K+ ion (mass of 6.5×10^−23kg) is accelerated through a cell membrane at 1.6 ×10^6 m/s2 What is the electric field between the membrane walls?50 pJ of energy is stored in a 2.0 cm x 2.0 cm x 2.0 cm region of uniform electric field. What is the electric field strength?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. 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: 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.
- A uniform electric field 3.41E3 N/C is applied in a cubic box with each side being 2.23 m long. What is the energy stored by the electric field in the box (in J)?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 ( ))A thundercloud produces a vertical electric field at a magnitude of 2.8 x10^4 N/C at ground level. You hold a 22.0 cm x 28.0 cm sheet of paper horizontally below the cloud. What is the electric flux through the sheet? An electric field of 500 V/m is destred between two parallel plates 153.0 mm apart. What voltoge should be applied?
- What is the magnitude, in microcoulombs per squared meter, of the surface charge density on the conducting plates? Given: Two parallel, flat, conducting plates with equal but opposite charges are separated by a uniform layer of insulating material with a dielectric constant of 7.1 and a thickness of 1.7mmmm. The electric field in the dielectric material is 2.22MV/mMV/m.(a) Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 10.0 g. Silver has 47 electrons per atom, and its molar mass is 107.87 g/mol. (b) Imagine adding electrons to the pin until the negative charge has the very large value 1.00 mC. How many electrons are added for every 109 electrons already present?(a) Find the total electric field at x = 1.00 cm in Figure 18.52(b) given that q =5.00 nC. (b) Find the total electric field at x = 11.00 cm in Figure 18.52(b). (c) If the charges are allowed to move and eventually be brought to rest by friction, what will the final charge configuration be? (That is, will there be a single charge, double charge; etc., and what will its value(s) he?)
