Q63 / 75 · JEE Main 2026 (06 April Shift 1) · Physics
A thin half ring of radius $35$ cm is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100$ V/m, then the value of $Q$ is _______ nC. ($\epsilon_o = 8.85 \times 10^{-12}$ C$^2$/Nm$^2$ and $\pi = 3.14$)
- Correct answerA$2.14$
- B$2.44$
- C$3.25$
- D$0.7$
Solution
The electric field at the center of a uniformly charged half ring is given by:
$E = \dfrac{2k\lambda}{R}$
where $\lambda = \dfrac{Q}{\pi R}$ is the linear charge density.
Substituting $\lambda$, we get:
$E = \dfrac{2kQ}{\pi R^2} = \dfrac{2Q}{4\pi\epsilon_0 \pi R^2} = \dfrac{Q}{2\pi^2\epsilon_0 R^2}$
Rearranging for $Q$:
$Q = 2\pi^2\epsilon_0 R^2 E$
Given values:
$E = 100$ V/m
$R = 35$ cm $= 0.35$ m
$\epsilon_0 = 8.85 \times 10^{-12}$ C$^2$/Nm$^2$
$\pi = 3.14$
Substituting the values:
$Q = 2 \times (3.14)^2 \times 8.85 \times 10^{-12} \times (0.35)^2 \times 100$
$Q = 2 \times 9.8596 \times 8.85 \times 10^{-12} \times 0.1225 \times 100$
$Q \approx 2137.8 \times 10^{-12}$ C
$Q \approx 2.14 \times 10^{-9}$ C $= 2.14$ nC
Answer: $2.14$
$E = \dfrac{2k\lambda}{R}$
where $\lambda = \dfrac{Q}{\pi R}$ is the linear charge density.
Substituting $\lambda$, we get:
$E = \dfrac{2kQ}{\pi R^2} = \dfrac{2Q}{4\pi\epsilon_0 \pi R^2} = \dfrac{Q}{2\pi^2\epsilon_0 R^2}$
Rearranging for $Q$:
$Q = 2\pi^2\epsilon_0 R^2 E$
Given values:
$E = 100$ V/m
$R = 35$ cm $= 0.35$ m
$\epsilon_0 = 8.85 \times 10^{-12}$ C$^2$/Nm$^2$
$\pi = 3.14$
Substituting the values:
$Q = 2 \times (3.14)^2 \times 8.85 \times 10^{-12} \times (0.35)^2 \times 100$
$Q = 2 \times 9.8596 \times 8.85 \times 10^{-12} \times 0.1225 \times 100$
$Q \approx 2137.8 \times 10^{-12}$ C
$Q \approx 2.14 \times 10^{-9}$ C $= 2.14$ nC
Answer: $2.14$
Hence, the correct answer is option: 'A'.