Question
Question: Eight dipoles of charges of magnitude q are placed inside a cube. Then, the total electric flux comi...
Eight dipoles of charges of magnitude q are placed inside a cube. Then, the total electric flux coming out of the cube will be
& A.\,\dfrac{8q}{\varepsilon } \\\ & B.\,\dfrac{16q}{\varepsilon } \\\ & C.\,\dfrac{q}{\varepsilon } \\\ & D.\,0 \\\ \end{aligned}$$Solution
The electric flux is the rate of flow of the electric field through a given area/surface. The dipole is a pair of positive and negative charges. The total electric flux coming out of the cube will be the sum of charges by epsilon.
Complete answer:
According to Gauss law, closed surfaces of various shapes can surround the charge. The net flux through closed surface equals ε01times the net charge enclosed by that surface. The product of the closed integral of the electric field and a small area is also the net electric flux. The mathematical representation of the same is given as follows.
Φ=∮E.dA=ε0qnet
From the statement, we can notice that the charges enclosed by the cube are 8 pairs of +q and –q. So, while computing the flux of the electric field through the surface S, we should consider only these charges.
Now, we will compute the total electric flux coming out of the cube.
The charge present on one dipole is, q=(+e)+(−e)∴q=0
So, the charge present on all the dipoles will also be zero.