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A long straight metallic pipe has radius a and surface charge density 6.A point located a distance 3a from the axis of the pipe has been designated as the zero of potential energy. + Part (a) Due to the cylindrical symmetry, the electric field may be expressed as E = E, where is perpendicular to the axis of the pipe. Enter an expression for the radial component of the electric field, E,, that is valid outside of the pipe, where r > a. E, = o a/(tor) expression E, = 0 Part (b) Due to the cylindrical symmetry, the electric field may be expressed as E = E, where is perpendicular to the axis of the pipe. Enter an for the radial component of the electric field, E,, that is valid outside of the pipe, where r < a. Part (e) Enter an expression for the electric potential, V, that is valid outside of the pipe, where r > a. Recall that a point located a distance 3a from the axis of the pipe has been designated as the zero of potential energy. V = (σ a)/E In(r/3a)

A long straight metallic pipe has radius a and surface charge density 6.A point located a distance 3a from the axis of the pipe has been designated as the zero of potential energy. + Part (a) Due to the cylindrical symmetry, the electric field may be expressed as E = E, where is perpendicular to the axis of the pipe. Enter an expression for the radial component of the electric field, E,, that is valid outside of the pipe, where r > a. E, = o a/(tor) expression E, = 0 Part (b) Due to the cylindrical symmetry, the electric field may be expressed as E = E, where is perpendicular to the axis of the pipe. Enter an for the radial component of the electric field, E,, that is valid outside of the pipe, where r < a. Part (e) Enter an expression for the electric potential, V, that is valid outside of the pipe, where r > a. Recall that a point located a distance 3a from the axis of the pipe has been designated as the zero of potential energy. V = (σ a)/E In(r/3a)

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solve parts c, d, and e please
A long straight metallic pipe has radius a and surface charge density o. A point located a distance 3a from the axis of the pipe has been designated as the zero of potential energy. + + + Part (a) Due to the cylindrical symmetry, the electric field may be expressed as E = E, where is perpendicular to the axis of the pipe. Enter an expression for the radial component of the electric field, E,, that is valid outside of the pipe, where r > a. E, = 0 a/(Eor) Part (d) Enter an expression for the electric potential, V, that is valid inside of the pipe, where r <a. V = o a/(to ln(3)) Part (b) Due to the cylindrical symmetry, the electric field may be expressed as E = E, where is perpendicular to the axis of the pipe. Enter an expression for the radial component of the electric field, Er, that is valid outside of the pipe, where r <a. E, = 0 Part (e) Enter an expression for the electric potential, V, that is valid outside of the pipe, where r > a. Recall that a point located a distance 3a from the axis of the pipe has been designated as the zero of potential energy. V = (σa)/Eo In(r/3a) Part (e) Designating a different point as the zero of electric potential is equivalent to adding a constant to the potential function. Suppose that you wanted to choose r = 2a instead of r = 3a as the zero of electric potential by adding a constant, Vshift, to your responses to step (c) and step (d). Enter an expression for Vshift- Vshift =
Transcribed Image Text:A long straight metallic pipe has radius a and surface charge density o. A point located a distance 3a from the axis of the pipe has been designated as the zero of potential energy. + + + Part (a) Due to the cylindrical symmetry, the electric field may be expressed as E = E, where is perpendicular to the axis of the pipe. Enter an expression for the radial component of the electric field, E,, that is valid outside of the pipe, where r > a. E, = 0 a/(Eor) Part (d) Enter an expression for the electric potential, V, that is valid inside of the pipe, where r <a. V = o a/(to ln(3)) Part (b) Due to the cylindrical symmetry, the electric field may be expressed as E = E, where is perpendicular to the axis of the pipe. Enter an expression for the radial component of the electric field, Er, that is valid outside of the pipe, where r <a. E, = 0 Part (e) Enter an expression for the electric potential, V, that is valid outside of the pipe, where r > a. Recall that a point located a distance 3a from the axis of the pipe has been designated as the zero of potential energy. V = (σa)/Eo In(r/3a) Part (e) Designating a different point as the zero of electric potential is equivalent to adding a constant to the potential function. Suppose that you wanted to choose r = 2a instead of r = 3a as the zero of electric potential by adding a constant, Vshift, to your responses to step (c) and step (d). Enter an expression for Vshift- Vshift =
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