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Physics Question on Electric Dipole

The electric flux for Gaussian surface A that enclose the charged particles in free space is (given q1=−14 nc,q2=78.85 nC,q3=−56 nCq_1= - 14\, nc, q_2 = 78.85\, nC, q_3 = - 56\, nC)

A

103 Nm2C−110^3\,Nm^2C^{-1}

B

103 CN−1m−210^3\,CN^{-1}m^{-2}

C

6.32×103 Nm2C−16.32 \times 10^3\,Nm^2C^{-1}

D

6.32×103 CN−1m−26.32 \times 10^3\,CN^{-1}m^{-2}

Answer

103 Nm2C−110^3\,Nm^2C^{-1}

Explanation

Solution

Electric flux =qε0=\frac{q}{\varepsilon_{0}} =(−14+7885−56)×10−9885×10−12=885×10−9885×10−12=\frac{(-14+7885-56) \times 10^{-9}}{885 \times 10^{-12}}=\frac{885 \times 10^{-9}}{885 \times 10^{-12}} =1000 Nm2 C−1=1000\, Nm ^{2} \,C ^{-1}