A ring with radius R and a uniformly distributed total charge Q lies in the xy plane, centered at the origin. Part A What is the potential V(z) due to the ring on the z axis as a function of z? Express your answer in terms of Q, z, R, and elementof_0 or k = 1/4 pi elementof_0. Part B What is the magnitude of the electric field E on the z axis as a function of z, for z > 0? Express your answer in terms of some or all of the quantities Q, z, R, and elementof_0 or k = By symmetry, the electric field has only one Cartesian component. In what direction does the electric field point? i j k

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06-12-2022
Physics

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A ring with radius R and a uniformly distributed total charge Q lies in the xy plane, centered at the origin. Part A What is the potential V(z) due to the ring on the z axis as a function of z? Express your answer in terms of Q, z, R, and elementof_0 or k = 1/4 pi elementof_0. Part B What is the magnitude of the electric field E on the z axis as a function of z, for z > 0? Express your answer in terms of some or all of the quantities Q, z, R, and elementof_0 or k = By symmetry, the electric field has only one Cartesian component. In what direction does the electric field point? i j k

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aamirmajeed5100 aamirmajeed5100 · Answer 1 · 2022-12-18T03:26:30-05:00

The potential V(z) due to the ring is given by the equation V(z) = (μQ/2πR)*cos(2πz/R).

where μ is the magnetic dipole moment of the ring, Q is the charge of the ring, R is the radius of the ring, and z is the distance from the center of the ring.

The electric field magnitude is calculated using the equation E = V/d, where V is the voltage and d is the distance between the two points in the field.

Part A

The potential V(z) due to the ring on the z-axis is given by:

V(z) = kQ/(2R) * ln(z + sqrt(R^2 + z^2))

where k = 1/4piε0 is Coulomb's constant.

Part B

The magnitude of the electric field E on the z-axis as a function of z, for z > 0, is given by:

E = kQ/(2R) * (1/sqrt(R^2 + z^2))

The electric field points away from the origin, in the positive z direction.

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