$\sigma$ is the uniform surface charge density of a thin spherical shell of radius $R$. The electric field at any point on the surface of the spherical shell is:

Question

A thin half ring of radius $35 \text{ cm}$ is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100 \text{ V/m}$, then the value of $Q$ is ______ $\text{nC}$.
($\epsilon_o = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$ and $\pi = 3.14$)

Answer 1

According to Gauss's Law:

\[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{in}}}{\varepsilon_0}. \]

For a spherical shell, the electric field $ E $ is uniform over the surface area:

\[ E \cdot 4\pi R^2 = \frac{\sigma \cdot 4\pi R^2}{\varepsilon_0}. \]

gaussin surface

\[ E = \frac{\sigma}{\varepsilon_0}. \]

Therefore, the electric field at the surface of the spherical shell is:

\[ \boxed{\frac{\sigma}{\varepsilon_0}}. \]

\(\sigma\) is the uniform surface charge density of a thin spherical shell of radius \(R\). The electric field at any point on the surface of the spherical shell is:

The Correct Option is C

Solution and Explanation

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