To find the electric field using Gauss's law: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q}{\varepsilon_0}, \] where \( Q \) is the total charge enclosed and \( \varepsilon_0 \) is the permittivity of free space.
Step 1: Relate charge to surface charge density. The total charge \( Q \) on the spherical shell is related to the surface charge density \( \sigma \) and the surface area of the shell (which is \( 4\pi R^2 \)): \[ Q = \sigma \cdot 4\pi R^2. \]
Step 2: Simplify the electric field. By symmetry, the electric field on the surface of the spherical shell is uniform, and the flux through the Gaussian surface (a sphere of radius \( R \)) is: \[ \oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi R^2, \] where \( E \) is the magnitude of the electric field. Substituting into Gauss's law: \[ E \cdot 4\pi R^2 = \frac{Q}{\varepsilon_0}. \] Substitute the expression for \( Q \): \[ E \cdot 4\pi R^2 = \frac{\sigma \cdot 4\pi R^2}{\varepsilon_0}. \] Canceling \( 4\pi R^2 \) from both sides: \[ E = \frac{\sigma}{\varepsilon_0}. \]
Final Answer: \[ \boxed{\frac{\sigma}{\varepsilon_0}}. \]
An infinite sheet of uniform charge \( \rho_s = 10\, {C/m}^2 \) is placed on the \( z = 0 \) plane. The medium surrounding the sheet has a relative permittivity of 10 . The electric flux density, in C/m\(^2\), at a point \( P(0, 0, 5) \), is: Note: \( \hat{a}, \hat{b}, \hat{c} \) are unit vectors along the \( x, y, z \) directions, respectively.
A metallic ring is uniformly charged as shown in the figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to O is ‘E’ magnitude. What would be the magnitude of electric field at ‘O’ due to arc ABC?
What is the empirical formula of a compound containing 40% sulfur and 60% oxygen by mass?