Answer :
Certainly! Let's solve the problem step-by-step.
### Step 1: Convert charges from microCoulombs (iC) to Coulombs (C)
Given charges:
- [tex]\( q_1 = 15 \, \mu\text{C} \)[/tex]
- [tex]\( q_2 = -17 \, \mu\text{C} \)[/tex]
Since [tex]\( 1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C} \)[/tex]:
[tex]\[ q_1 = 15 \times 10^{-6} \, \text{C} = 15 \times 10^{-6} \, \text{C} \][/tex]
[tex]\[ q_2 = -17 \times 10^{-6} \, \text{C} = -17 \times 10^{-6} \, \text{C} \][/tex]
### Step 2: Calculate the total charge enclosed by the surface
The total charge enclosed, [tex]\( q_{\text{total}} \)[/tex], is the sum of both charges:
[tex]\[ q_{\text{total}} = q_1 + q_2 = 15 \times 10^{-6} \, \text{C} + (-17 \times 10^{-6} \, \text{C}) \][/tex]
[tex]\[ q_{\text{total}} = (15 - 17) \times 10^{-6} \, \text{C} = -2 \times 10^{-6} \, \text{C} \][/tex]
### Step 3: Use Gauss's Law to calculate the total electric flux
Gauss's Law states that the total electric flux, [tex]\( \Phi_E \)[/tex], through a closed surface is equal to the total charge enclosed divided by the electric constant, [tex]\( \epsilon_0 \)[/tex]:
[tex]\[ \Phi_E = \frac{q_{\text{total}}}{\epsilon_0} \][/tex]
The electric constant [tex]\( \epsilon_0 \)[/tex] is approximately [tex]\( 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \)[/tex]:
[tex]\[ \Phi_E = \frac{-2 \times 10^{-6} \, \text{C}}{8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)} \][/tex]
### Step 4: Calculate the numerical value of the flux
[tex]\[ \Phi_E = -2 \times 10^{-6} \, \text{C} \div 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \][/tex]
Performing the division:
[tex]\[ \Phi_E \approx -225886.60492432796 \, \text{N} \cdot \text{m}^2/\text{C} \][/tex]
### Step 5: Convert the flux into the desired units (Nm²/C)
To match the comparable form, we convert the flux to [tex]\( Nm²C^{-1} \)[/tex] by multiplying by [tex]\( 10^{-4} \)[/tex]:
[tex]\[ \Phi_{\text{comparable}} = -225886.60492432796 \times 10^{-4} \, \text{Nm}^2/\text{C} = -22.588660492432798 \, \text{Nm}^2/\text{C} \][/tex]
### Conclusion
Given the options provided:
a. 1.321
b. 6.284
c. 1.321
d. 0
None of the options directly match the calculated value of [tex]\( -22.588660492432798 \, \text{Nm}^2 \text{C}^{-1} \)[/tex] so it seems there might be a mistake in the given options or further context is required to match the calculated result.
### Step 1: Convert charges from microCoulombs (iC) to Coulombs (C)
Given charges:
- [tex]\( q_1 = 15 \, \mu\text{C} \)[/tex]
- [tex]\( q_2 = -17 \, \mu\text{C} \)[/tex]
Since [tex]\( 1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C} \)[/tex]:
[tex]\[ q_1 = 15 \times 10^{-6} \, \text{C} = 15 \times 10^{-6} \, \text{C} \][/tex]
[tex]\[ q_2 = -17 \times 10^{-6} \, \text{C} = -17 \times 10^{-6} \, \text{C} \][/tex]
### Step 2: Calculate the total charge enclosed by the surface
The total charge enclosed, [tex]\( q_{\text{total}} \)[/tex], is the sum of both charges:
[tex]\[ q_{\text{total}} = q_1 + q_2 = 15 \times 10^{-6} \, \text{C} + (-17 \times 10^{-6} \, \text{C}) \][/tex]
[tex]\[ q_{\text{total}} = (15 - 17) \times 10^{-6} \, \text{C} = -2 \times 10^{-6} \, \text{C} \][/tex]
### Step 3: Use Gauss's Law to calculate the total electric flux
Gauss's Law states that the total electric flux, [tex]\( \Phi_E \)[/tex], through a closed surface is equal to the total charge enclosed divided by the electric constant, [tex]\( \epsilon_0 \)[/tex]:
[tex]\[ \Phi_E = \frac{q_{\text{total}}}{\epsilon_0} \][/tex]
The electric constant [tex]\( \epsilon_0 \)[/tex] is approximately [tex]\( 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \)[/tex]:
[tex]\[ \Phi_E = \frac{-2 \times 10^{-6} \, \text{C}}{8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)} \][/tex]
### Step 4: Calculate the numerical value of the flux
[tex]\[ \Phi_E = -2 \times 10^{-6} \, \text{C} \div 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \][/tex]
Performing the division:
[tex]\[ \Phi_E \approx -225886.60492432796 \, \text{N} \cdot \text{m}^2/\text{C} \][/tex]
### Step 5: Convert the flux into the desired units (Nm²/C)
To match the comparable form, we convert the flux to [tex]\( Nm²C^{-1} \)[/tex] by multiplying by [tex]\( 10^{-4} \)[/tex]:
[tex]\[ \Phi_{\text{comparable}} = -225886.60492432796 \times 10^{-4} \, \text{Nm}^2/\text{C} = -22.588660492432798 \, \text{Nm}^2/\text{C} \][/tex]
### Conclusion
Given the options provided:
a. 1.321
b. 6.284
c. 1.321
d. 0
None of the options directly match the calculated value of [tex]\( -22.588660492432798 \, \text{Nm}^2 \text{C}^{-1} \)[/tex] so it seems there might be a mistake in the given options or further context is required to match the calculated result.