Question

Electric field at the centre of a semi-circular ring of radius \(10\,\text{cm}\) is \(100\,\text{V/m}\). Find the charge on the ring if the charge distribution is uniform.

• A. \(4\varepsilon_0\)
• B. \(20\varepsilon_0\)
• C. \(25\varepsilon_0\)
• D. \(30\varepsilon_0\)

Answer 1

The Correct Option is B

Concept: For a uniformly charged semicircular ring, the horizontal components of electric field cancel due to symmetry, while the vertical components add. The electric field at the centre of a uniformly charged semicircular ring is \[ E = \frac{2k\lambda}{R} \] where

• \(k = \dfrac{1}{4\pi\varepsilon_0}\)
• \(\lambda\) = linear charge density
• \(R\) = radius of the semicircle

The linear charge density is \[ \lambda = \frac{Q}{\pi R} \] Substituting this into the electric field formula gives \[ E = \frac{2kQ}{\pi R^2} \] Step 1: Substitute the known values.} Given \[ E = 100 \, \text{V/m}, \quad R = 10\,\text{cm} = 0.1\,\text{m} \] Using \[ E = \frac{2kQ}{\pi R^2} \] \[ 100 = \frac{2}{\pi (0.1)^2}\left(\frac{1}{4\pi\varepsilon_0}\right)Q \] Step 2: Solve for total charge \(Q\).} Rearranging, \[ Q = \frac{100 \times \pi \times (0.1)^2}{2k} \] Substitute \(k = \dfrac{1}{4\pi\varepsilon_0}\) \[ Q = \frac{100 \times \pi \times 0.01}{2} \times 4\pi\varepsilon_0 \] \[ Q = 2\pi^2 \varepsilon_0 \] Since \(\pi^2 \approx 10\), \[ Q \approx 20\varepsilon_0 \] Thus, \[ \boxed{Q = 20\varepsilon_0} \]