Question

Two point charges q = 63.0 nC and q = ?63.0 nC are in vacuum located on the x axis at ?46.0 cm and 46.0 cm, respectively. Choosing the normal to the plane in the positive x direction, calculate the total electric flux through the y?z plane.

Answer 1

SOLUTION

The positive charge is placed in the negative x-axis and the negative charge is placed in the positive x-axis. Additionally the normal $\hat{u}_{n}$ to the y-z plane points in the positive x-direction, so we have:

Now, the electric flux $\Phi_{E}$ is defined as:

$\Phi_{E}=\frac{q}{\varepsilon_{o}}$

But, due to y-z plane is infinite, and charge are very near him, the electric flux of each charge is half, so:

$\Phi_{1}=\Phi_{2}=\frac{q}{2\varepsilon_{o}}$

Now, according to figure given above, the electric field lines of both charges points at the same direction, therefore the total electric flux $\Phi_{t}$ through the y-z plane is:

$\Phi_{t}=\Phi_{1}+\Phi_{2}=\frac{q}{2\varepsilon_{o}}+\frac{q}{2\varepsilon_{o}}={\color{Red} \frac{q}{\varepsilon_{o}}}$

Replacing values we obtain:

$\Phi_{t}= \frac{q}{\varepsilon_{o}}=\frac{63\times 10^{-9}\,C}{8.85\times 10^{-12}C^{2}/N.m^{2}}={\color{Blue} 7.12\times 10^{3}\,\frac{N.m^{2}}{C}}$