Question

A sphere encloses charges \(+5\,C\) and \(-2\,C\), while a charge \(-3\,C\) is outside the sphere. What is the electric flux through the sphere?

• A. \(\dfrac{3C}{\varepsilon_{0}}\)
• B. \(-\dfrac{3C}{\varepsilon_{0}}\)
• C. Zero
• D. \(\dfrac{10C}{\varepsilon_{0}}\)

Hint:

For any closed Gaussian surface, \[ \boxed{\Phi=\frac{Q_{\text{enclosed}}}{\varepsilon_{0}}} \] Always calculate the net enclosed charge only. Remember: \[ \boxed{ \begin{aligned} \text{Positive enclosed charge} &\Rightarrow \text{Positive flux},\\ \text{Negative enclosed charge} &\Rightarrow \text{Negative flux},\\ \text{External charges} &\Rightarrow \text{No effect on net flux}. \end{aligned} } \]

Answer 1

The Correct Option is A

Concept: The electric flux through any closed surface is determined by Gauss's Law. According to Gauss's law, \[ \boxed{\Phi=\oint \vec{E}\cdot d\vec{A}=\frac{Q_{\text{enclosed}}}{\varepsilon_{0}}} \] where \[ \Phi=\text{Electric flux through the closed surface}, \] \[ Q_{\text{enclosed}}=\text{Net charge enclosed inside the surface}, \] \[ \varepsilon_{0}=\text{Permittivity of free space}. \] A very important point is that only the charges enclosed within the closed surface contribute to the electric flux. Any charge present outside the surface produces equal numbers of field lines entering and leaving the surface, resulting in zero net contribution.

Step 1: Determine the net charge enclosed by the sphere.
The sphere encloses the charges \[ +5\,C \] and \[ -2\,C. \] Hence, \[ Q_{\text{enclosed}} = (+5)+(-2) = 3\,C. \] Therefore, \[ \boxed{Q_{\text{enclosed}}=3\,C.} \]

Step 2: Consider the charge outside the sphere.
The charge \[ -3\,C \] is located outside the Gaussian surface. According to Gauss's law, an external charge does not affect the net electric flux through a closed surface. Hence, \[ \boxed{\text{The outside charge contributes zero net flux.}} \]

Step 3: Apply Gauss's law.
Using \[ \Phi=\frac{Q_{\text{enclosed}}}{\varepsilon_{0}}, \] we obtain \[ \Phi = \frac{3}{\varepsilon_{0}}. \] Therefore, \[ \boxed{\Phi=\frac{3C}{\varepsilon_{0}}.} \] Hence, the correct answer is \[ \boxed{\textbf{Option (A)}}. \]

Verification: Even though the total charge in the system is \[ (+5)+(-2)+(-3)=0, \] Gauss's law depends only on the charge enclosed by the Gaussian surface, not on the total charge present in space. Since the enclosed charge is \[ 3\,C, \] the electric flux remains \[ \boxed{\dfrac{3C}{\varepsilon_{0}}.} \]