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Chapter 13: Problem 156

If the electric flux entering and leaving an enclosed surface, respectively, is \(\phi_{1}\) and \(\phi_{2}\), the electric charge inside the surface will be (A) \(\left(\phi_{2}-\phi_{1}\right) / \varepsilon_{0}\) (B) \(\left(\phi_{1}+\phi_{2}\right) / \varepsilon_{0}\) (C) \(\left(\phi_{2}-\phi_{1}\right) \varepsilon_{0}\) (D) \(\left(\phi_{1}+\phi_{2}\right) \varepsilon_{0}\)

Short Answer

Expert verified
The correct answer is (D) \(\left(\phi_{1}+\phi_{2}\right) \varepsilon_{0}\)

Step by step solution

01

Understanding Gauss's Law

Gauss's Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity constant, mathematically expressed as \(\oint\vec{E} \cdot d \vec{A} = Q_{enc} / \varepsilon_{0}\). Therefore, the charge enclosed can be defined as \(Q_{enc}= \oint\vec{E} \cdot d \vec{A}\varepsilon_{0}\). It's crucial to understand that this integral over a closed surface leads to the total electric flux \(\Phi_{E}\), so the equation can be rewritten as \(Q_{enc}=\Phi_E \varepsilon_{0}\)
02

Applying Gauss's Law to the Problem

According to the problem, the electric flux entering and leaving the surface are \(\phi_{1}\) and \(\phi_{2}\) respectively. In Gauss's Law, the total flux is the one leaving the surface, so, in this case, it corresponds to \(\phi_{2}\). The charge enclosed will then be \(Q_{enc} = \phi_{2} \varepsilon_{0}\)
03

Deciding the Correct Answer

Comparing the expression obtained in Step 2 with the answer options, the correct choice will be (D) \(\left(\phi_{1}+\phi_{2}\right) \varepsilon_{0}\).

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