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A load Q = -820 nC is uniformly distributed in a ring of radius 2.4 m. A point load q = +530 nC is stationary in the centre of the ring. Points A and B lie on the axis of the ring. What is the minimum work that an external force must produce to transfer the electron from B to A? (e = 1,60 × 10-19 C, k = 1/4pe0 = 8,99 × 109 N∙m2/C2)

User Ssierral
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To calculate the minimum work required to move the point charge q from point B to A, we need to calculate the electric potential difference between the two points, and then use the equation W = qΔV, where W is the work done, q is the charge being moved, and ΔV is the potential difference.

To find the electric potential at point A and B due to the charged ring, we can use the equation for electric potential due to a charged ring:

V = kQ/r

Where k is Coulomb's constant, Q is the total charge of the ring, and r is the distance from the center of the ring to the point where the potential is being calculated.

For point B, the potential due to the charged ring is:

VB = kQ/r = (8.99 × 10^9 N·m^2/C^2) * (-820 × 10^-9 C) / (2.4 m) = -306.55 V

For point A, the potential due to the charged ring is:

VA = kQ/r = (8.99 × 10^9 N·m^2/C^2) * (-820 × 10^-9 C) / (4.8 m) = -153.27 V

The potential difference between point A and B is:

ΔV = VA - VB = (-153.27 V) - (-306.55 V) = 153.28 V

The minimum work required to move the charge q from point B to A is:

W = qΔV = (530 × 10^-9 C) * (153.28 V) = 81.09 × 10^-6 J

Therefore, the minimum work required to transfer the electron from B to A is 81.09 × 10^-6 J.

User Estevao Santiago
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