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Electric flux connects the geometry of conductors to the fields they generate. Learn this powerful tool and shortcut your way to the electric field of symmetrical arrangements like wires and sheets.

If the total charge on the surface of a conducting sphere is \(2.3 \,\mu\text{C}\) and the diameter of the sphere is \(1.1\text{ m},\) what is the magnitude of the electric field just outside the surface of the sphere, due to the surface charge?

The value of the permittivity constant is \(\varepsilon_0=8.85 \times 10^{-12} \text{ C}^2\text{/N}\cdot\text{m}^2.\)

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Suppose that a charged particle is held at the center of two concentric conducting spherical shells \(A\) and \(B,\) as shown in the above figure. The radius of sphere \(A\) is \(r_A=2\text{ cm}\) and that of sphere \(B\) is \(r_B=4\text{ cm}.\) The net flux \(\Phi\) through a Gaussian sphere centered on the particle is
\[\begin{align}
-16.0 \times 10^5 \text{ N}\cdot\text{m}^2\text{/C} &\text{for } 0 < r < r_A \\
+8.0 \times 10^5 \text{ N}\cdot\text{m}^2\text{/C} &\text{for } r_A < r < r_B\\
-4.0 \times 10^5 \text{ N}\cdot\text{m}^2\text{/C} &\text{for } r > r_A,
\end{align}\] where \(r\) is the radius of the Gaussian sphere. Then what are the net charges of shell \(A\) and shell \(B,\) respectively?

The value of the permittivity constant is \(\varepsilon_0=8.85 \times 10^{-12} \text{ C}^2\text{/N}\cdot\text{m}^2.\)

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Consider a uniformly charged conducting sphere. If the radius of the sphere is \(0.50\text{ m}\) and the surface charge density is \(8.40 \,\mu\text{C/m}^2,\) what is the approximate total electric flux leaving the surface of the sphere?

The value of the permittivity constant is \(\varepsilon_0=8.85 \times 10^{-12} \text{ C}^2\text{/N}\cdot\text{m}^2.\)

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If the magnitude of electric field just above the surface of a charged conducting cylinder is \(2.8 \times 10^5 \text{ N/C},\) what is the surface charge density of the cylinder?

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