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Chapter 22: Problem 55

A thin ring of radius \(R\) carries charge \(3 Q\) distributed uniformly over three-fourths of its circumference, and \(-Q\) over the rest. Find the potential at the ring's center.

Short Answer

Expert verified
The electric potential at the ring's center is \(V = k \frac{2Q}{R}\).

Step by step solution

01

Calculate the potential due to the charge \(3Q\)

Considering the \(3Q\) charge distributed uniformly over three-fourths of the ring's circumference, the potential \(\(V_{3Q}\)\) at the center is calculated using the formula for potential due to a ring at its center, \(V = k \frac{Q}{r}\). In this case, \(Q = 3Q\) and \(r = R\), hence \(V_{3Q} = k \frac{3Q}{R}\).
02

Calculate the potential due to the charge \(-Q\)

Next, considering the \(-Q\) charge distributed over the remaining one-fourth of the ring's circumference, the potential \(\(V_{-Q}\)\) at the center is also calculated using the formula for potential due to a ring at its center, \(V = k \frac{Q}{r}\). Here, \(Q = -Q\) and \(r = R\), hence \(V_{-Q} = k \frac{-Q}{R}\).
03

Add the potentials

Finally, the total potential \(V\) at the center of the thin ring with this non-uniform charge distribution is obtained by adding the potential due to the \(3Q\) charge to the potential due to the \(-Q\) charge, i.e. \(V = V_{3Q} + V_{-Q}\). Substituting the previously calculated values, \(V = k \frac{3Q}{R} + k \frac{-Q}{R} = k \frac{2Q}{R}\).

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