Question

charge densities of the four quadrants of the ring are ( lambda, -2lambda, -lambda ) and ( +2lambda ) respectively. the electric potential at the centre of ring is

1. ( \frac{sqrt{2}lambda}{...} )
2. zero
3. ( \frac{sqrt{2}lambda}{...} )
4. ( \frac{sqrt{5}lambda}{...} )

Explanation:

Step1: Recall Electric Potential Formula

The electric potential at the center of a charged ring segment with linear charge density \(\lambda\), length \(L\), and radius \(R\) is given by \(V = \frac{kQ}{R}\), where \(Q=\lambda L\) and \(k=\frac{1}{4\pi\epsilon_0}\). For a quadrant, the length \(L=\frac{\pi R}{2}\).

Step2: Calculate Potential for Each Quadrant

• For \(\lambda\) (first quadrant): \(Q_1=\lambda\times\frac{\pi R}{2}\), \(V_1=\frac{kQ_1}{R}=\frac{k\lambda\pi R/2}{R}=\frac{k\lambda\pi}{2}\)
• For \(-2\lambda\) (second quadrant): \(Q_2=-2\lambda\times\frac{\pi R}{2}\), \(V_2=\frac{kQ_2}{R}=\frac{k(-2\lambda)\pi R/2}{R}=-k\lambda\pi\)
• For \(-\lambda\) (third quadrant): \(Q_3=-\lambda\times\frac{\pi R}{2}\), \(V_3=\frac{kQ_3}{R}=\frac{k(-\lambda)\pi R/2}{R}=-\frac{k\lambda\pi}{2}\)
• For \(+2\lambda\) (fourth quadrant): \(Q_4=2\lambda\times\frac{\pi R}{2}\), \(V_4=\frac{kQ_4}{R}=\frac{k(2\lambda)\pi R/2}{R}=k\lambda\pi\)

Step3: Sum the Potentials

Total potential \(V = V_1 + V_2 + V_3 + V_4\)
Substitute values:
\(V=\frac{k\lambda\pi}{2}-k\lambda\pi-\frac{k\lambda\pi}{2}+k\lambda\pi\)
Simplify term by term:
\(\frac{k\lambda\pi}{2}-\frac{k\lambda\pi}{2}=0\) and \(-k\lambda\pi + k\lambda\pi = 0\), so \(V = 0\)

Answer:

1. Zero