Gauss's law and Coulomb's law , although expressed in different forms , are equivalent ways of describing the relation between charge and electric field in static conditions . Gauss's law is `epsilon_(0) phi = q_(encl)`,when `q(encl)` is the net charge inside an imaginary closed surface called Gaussian surface. The two equations hold only when the net charge is in vaccum or air .
The net flux of the electric field through the surface due to `q_(3)` and `q_(4)` is

Options

A. zero
B. positive
C. negative
D. cannot say

Answer

To solve the problem of finding the net electric flux through a Gaussian surface due to charges \( q_3 \) and \( q_4 \), we can follow these steps:

### Step 1: Understand Gauss's Law
Gauss's Law states that the electric flux \( \Phi_E \) through a closed surface is proportional to the charge enclosed \( q_{encl} \) within that surface. Mathematically, it is given by:

\[ \Phi_E = \frac{q_{encl}}{\epsilon_0} \]

where \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Identify the Charges
Assume we have two point charges \( q_3 \) and \( q_4 \). We need to determine whether these charges are inside or outside the Gaussian surface we are considering.

### Step 3: Calculate the Enclosed Charge
If both charges \( q_3 \) and \( q_4 \) are inside the Gaussian surface, then the total enclosed charge \( q_{encl} \) is:

\[ q_{encl} = q_3 + q_4 \]

If one or both charges are outside the surface, then they do not contribute to the enclosed charge, and we only sum the charges that are inside.

### Step 4: Apply Gauss's Law
Using Gauss's Law, we can now calculate the electric flux through the surface:

\[ \Phi_E = \frac{q_{encl}}{\epsilon_0} \]

### Step 5: Substitute Values
If we know the values of \( q_3 \) and \( q_4 \), we can substitute them into the equation to find the net electric flux.

### Example Calculation
Assuming \( q_3 = +2 \, \mu C \) and \( q_4 = -3 \, \mu C \):

1. Calculate the enclosed charge:
\[ q_{encl} = 2 \, \mu C - 3 \, \mu C = -1 \, \mu C \]

2. Calculate the electric flux:
\[ \Phi_E = \frac{-1 \times 10^{-6}}{8.85 \times 10^{-12}} \approx -113,000 \, \text{N m}^2/\text{C} \]

### Final Result
The net electric flux through the Gaussian surface due to charges \( q_3 \) and \( q_4 \) is approximately \( -113,000 \, \text{N m}^2/\text{C} \).

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