A circular wire of radius R carries a total charge Q distributed uniformly over its circumference. A small length of wire subtending angle theta at the centre is cut off. Find the magnitude of electric field at the centre due to the remaining portion?

Question

Answer

Problem Statement: Find the magnitude of electric field at the centre of a circular wire of radius R carrying total charge Q distributed uniformly over its circumference, after a small length of wire subtending angle theta at the centre is cut off.

Solution:

Step 1: Calculation of Charge Density

The charge density of the circular wire can be calculated as follows:

Charge density (σ) = Total charge (Q) / Circumference of the wire (2πR)

σ = Q / 2πR

Step 2: Calculation of Electric Field Due to the Full Circular Wire

The electric field at the center of the full circular wire can be calculated as follows:

Electric field (E) = σ / (2ε₀)

where ε₀ is the permittivity of free space.

Therefore, Electric field (E) = Q / (4πε₀R)

Step 3: Calculation of Electric Field Due to the Cut-Off Section

The electric field at the center due to the cut-off section can be calculated as follows:

Electric field (dE) = (σ / (4πε₀)) * (dθ / 2)

where dθ is the angle subtended by the cut-off section at the center.

Step 4: Calculation of Electric Field Due to the Remaining Circular Wire

The electric field at the center due to the remaining portion of the circular wire can be calculated as follows:

Electric field (E') = E - dE

Electric field (E') = Q / (4πε₀R) - (σ / (4πε₀)) * (dθ / 2)

Electric field (E') = Q / (4πε₀R) - (Q / (8π²ε₀R)) * (dθ / π)

Electric field (E') = Q / (4πε₀R) * (1 - (dθ / 2π))

Step 5: Final Answer

Therefore, the magnitude of electric field at the center due to the remaining portion of the circular wire is given by:

|E'| = Q / (4πε₀R) * (1 - (dθ / 2π))

Answer

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