Question

Five charges +q, +5q, -2q, +3q and -4q are situated as shown in the figure. The electric flux due to this configuration through the surface S is: 
 

• A. 5q/ε0
• B. 4q/ε0
• C. 3q/ε0
• D. q/ε0

Hint:

Remember Gauss's Law: The total electric flux through a closed surface is proportional to the net electric charge enclosed within the surface. Charges outside the surface do not contribute to the net flux through the surface.

Answer 1

The Correct Option is B

Using Gauss's law, the electric flux \( \phi \) is given by the formula: \[ \phi = \frac{q}{\epsilon_0} \] where: - \( q \) is the charge inside the closed surface, - \( \epsilon_0 \) is the permittivity of free space. Now, if there are multiple charges inside the surface, we sum up the individual charges. Here, we are given the charges inside the closed surface: \( q \), \( -2q \), and \( 5q \). So, the total charge \( q_{{total}} \) inside the surface is: \[ q_{{total}} = q + (-2q) + 5q \] \[ q_{{total}} = 4q \] Therefore, the electric flux is: \[ \phi = \frac{q_{{total}}}{\epsilon_0} = \frac{4q}{\epsilon_0} \] Thus, the electric flux \( \phi \) through the closed surface is: \[ \phi = \frac{4q}{\epsilon_0} \]

Answer 2

Step 1: Use Gauss’s Law
According to Gauss's law, the total electric flux \( \Phi \) through a closed surface is given by:
\[ \Phi = \frac{q_{\text{in}}}{\varepsilon_0} \]
where \( q_{\text{in}} \) is the net charge enclosed within the surface \( S \), and \( \varepsilon_0 \) is the permittivity of free space.

Step 2: Identify the charges inside the surface
From the diagram, the charges inside surface \( S \) are:
- \( +5q \)
- \( +q \)
- \( -2q \)

Step 3: Add the enclosed charges
\[ q_{\text{in}} = (+5q) + (+q) + (-2q) = 4q \]

Step 4: Apply Gauss’s Law
\[ \Phi = \frac{q_{\text{in}}}{\varepsilon_0} = \frac{4q}{\varepsilon_0} \]

Final Answer: \( \frac{4q}{\varepsilon_0} \)