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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.

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 ∊₀.

User Eliot Ball
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Final answer:

The potential V(z) due to a ring with uniformly distributed total charge Q on the z-axis can be calculated using the formula V(z) = (Q/4πε₀)×(1/√(z² + R²)).

Step-by-step explanation:

The potential V(z) due to the ring on the z-axis can be determined by integrating the electric potential dVp for each infinitesimal charge element on the ring. The expression for dVp is given by:

dVp = (1/4πε₀)×(dq/√(z² + r²))

where r is the radius of the ring and dq is the charge of the infinitesimal element. The total potential V(z) on the z-axis is obtained by integrating dVp over the entire ring:

V(z) = (Q/4πε₀)×(1/√(z² + R²))

where Q is the total charge on the ring. Therefore, the potential V(z) as a function of z is given by V(z) = (Q/4πε₀)×(1/√(z² + R²)).

The electric potential V(z) due to a charged ring on the z axis can be calculated using the principles of electrostatics. Given a ring with radius R and a uniformly distributed total charge Q lying in the xy plane, we find the potential at a distance z along the z-axis passing through the center of the ring. Due to symmetry, each infinitesimal charge element contributes equally to the potential at point P on the z axis.

The electric potential due to an infinitesimal charge element is given by the expression dV = (kdq)/(sqrt(z2 + R2)), where k is Coulomb's constant, and dq is the charge of the infinitesimal element. By integrating over the entire ring, we obtain the total potential as V(z) = (kQ)/(sqrt(z2 + R2)). Here, the Coulomb's constant k can be expressed in terms of the vacuum permittivity ε0 as k = 1/(4πε0). Therefore, V(z) = (Q)/(4πε0sqrt(z2 + R2)), which is the potential due to the charged ring at point P on the z axis.

User Sam Stephenson
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