Question:
A wire of length \(L\) has a charge \(Q\) distributed uniformly along its length. The wire is bent in the shape of a semicircle. The magnitude of the electric field at the centre of curvature of the semicircle is
A wire of length \(L\) has a charge \(Q\) distributed uniformly along its length. The wire is bent in the shape of a semicircle. The magnitude of the electric field at the centre of curvature of the semicircle is
Show Hint
For a uniformly charged semicircular arc,
\[
E=\frac{2k\lambda}{R}
\]
at the centre, because symmetric horizontal components cancel and vertical components add.
Updated On: Jun 25, 2026
- \(\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{L^2}\)
- \(\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q^2}{L}\)
- \(\dfrac{Q}{2\varepsilon_0L^2}\)
- \(\dfrac{1}{2\pi\varepsilon_0}\dfrac{Q}{L^2}\)
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The Correct Option is C
Solution and Explanation
Step 1: Find the radius of the semicircle.
Length of the wire equals the arc length of a semicircle: \[ L=\pi R \] Therefore, \[ R=\frac{L}{\pi} \]
Step 2: Determine the linear charge density.
Uniform linear charge density: \[ \lambda=\frac{Q}{L} \]
Step 3: Electric field due to a semicircular charged wire.
For a uniformly charged semicircular arc, horizontal components of electric field cancel due to symmetry, while vertical components add.
The electric field at the centre is \[ E=\frac{2k\lambda}{R} \] where \[ k=\frac{1}{4\pi\varepsilon_0} \] Substituting, \[ E=\frac{2}{R}\left(\frac{1}{4\pi\varepsilon_0}\right)\lambda \] \[ E=\frac{\lambda}{2\pi\varepsilon_0R} \]
Step 4: Substitute \(\lambda\) and \(R\).
Using \[ \lambda=\frac{Q}{L} \] and \[ R=\frac{L}{\pi}, \] we get \[ E= \frac{1}{2\pi\varepsilon_0} \cdot \frac{Q}{L} \cdot \frac{\pi}{L} \] \[ E=\frac{Q}{2\varepsilon_0L^2} \]
Step 5: Final conclusion.
Hence, the magnitude of the electric field at the centre is \[ \boxed{\frac{Q}{2\varepsilon_0L^2}} \]
Length of the wire equals the arc length of a semicircle: \[ L=\pi R \] Therefore, \[ R=\frac{L}{\pi} \]
Step 2: Determine the linear charge density.
Uniform linear charge density: \[ \lambda=\frac{Q}{L} \]
Step 3: Electric field due to a semicircular charged wire.
For a uniformly charged semicircular arc, horizontal components of electric field cancel due to symmetry, while vertical components add.
The electric field at the centre is \[ E=\frac{2k\lambda}{R} \] where \[ k=\frac{1}{4\pi\varepsilon_0} \] Substituting, \[ E=\frac{2}{R}\left(\frac{1}{4\pi\varepsilon_0}\right)\lambda \] \[ E=\frac{\lambda}{2\pi\varepsilon_0R} \]
Step 4: Substitute \(\lambda\) and \(R\).
Using \[ \lambda=\frac{Q}{L} \] and \[ R=\frac{L}{\pi}, \] we get \[ E= \frac{1}{2\pi\varepsilon_0} \cdot \frac{Q}{L} \cdot \frac{\pi}{L} \] \[ E=\frac{Q}{2\varepsilon_0L^2} \]
Step 5: Final conclusion.
Hence, the magnitude of the electric field at the centre is \[ \boxed{\frac{Q}{2\varepsilon_0L^2}} \]
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