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Physics Question on Electric charges and fields

The electric flux for Gaussian surface AA that encloses the charged particles in free is (Given q1,=−14 n C,q2=78.85 n C, q3=−56 n C)q_1, = - 14 \, n \, C, q_2 = 78.85 \, n \, C, \, q_3 = - 56 \, n \, C)

A

103 Nm2 C−1 10^3 \, Nm^2 \, C^{ - 1}

B

103 CN−1 C−2 10^3 \, CN^{ - 1} \, C^{ - 2}

C

6.32×103 Nm2 C−1 6.32 \times 10^3 \, Nm^2 \, C^{ - 1}

D

6.32×103 CN−1 m−2 6.32 \times 10^3 \, CN^{ - 1} \, m^{ - 2}

Answer

103 Nm2 C−1 10^3 \, Nm^2 \, C^{ - 1}

Explanation

Solution

Flux is due to charges enclosed per ε0\varepsilon_0 ∴\therefore Total flux= ( −14×78.85−56)nC/ε0- 14 \times 78.85 - 56 ) nC/ \varepsilon_0 = 8.85×10−9 C×4π4πε08.85 \times 10^{ - 9 } \, C \times \frac{ 4 \pi}{ 4 \pi \varepsilon_0 } = 8.85×10−9×9×109×4π 8.85 \times 10^{ - 9 } \times 9 \times 10^9 \times 4 \pi = 1000.4 Nm2/C ie 1000Nm2 C−1Nm^2/C \, ie \, 1000 Nm^2 \, C^{ - 1}