Exams › JEE Main › Physics

A uniformly charged semicircular wire has radius a and linear charge density λ. The magnitude of the electric field at the centre of the semicircle is

1. λ/(2π ε₀ a)
2. λ/(2π² ε₀ a)
3. λ/(4π² ε₀ a)
4. λ²/(2π ε₀ a)

Correct answer: λ/(2π ε₀ a)

Solution

The electric field at the center of a uniformly charged semicircular wire can be derived by integrating the contributions from each infinitesimal charge element along the semicircle. Due to symmetry, the horizontal components of the electric field cancel out, while the vertical components add up, resulting in a total electric field magnitude of λ/(2π ε₀ a).

Related JEE Main Physics questions

• What is the SI unit of the permittivity of free space, ε₀?
• What is the SI unit used for electric flux?
• A constant electric field E is directed along the positive x-axis. If a charge of 0.5 C is displaced by 2 m along a line inclined at 60° to the x-axis, and the work done in this process is 10 J, what is the magnitude of the electric field?
• An electric dipole is located at the origin O along the x-axis. A point P lies 20 cm from O, and the line OP makes an angle π/3 with the x-axis. If the electric field at P is inclined at an angle θ to the x-axis, what is θ?
• A point Q is located on the perpendicular bisector of an electric dipole whose dipole moment is p. If Q is at a distance r from the dipole and r is much greater than the dipole size, the electric field at Q varies as
• A solid sphere of radius R carries charge q spread throughout its volume with volume charge density given by ρ = k r^a, where k and a are constants and r is measured from the centre. If the electric field at the point r = R/2 is one-eighth of the field at the surface r = R, determine the value of a.

⚔️ Practice JEE Main Physics free + battle 1v1 →