# An electric field given by 1

An electric field given by $\vec E=4.0\vec i -3(y^2+2)\vec j$ pierces a Gaussian cube of edge length 2.0 m and positioned as shown. (The magnitude E is in newtons per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube? Image from: J. Walker, D. Halliday, and R. Resnick, Fundamentals of physics: [extended], 10th ed. United States: Wiley, John & Sons, 2013.

#### Solution:

a) Top face

$\Phi= \displaystyle\oint \vec E \cdot d\vec A$

$\Phi= \displaystyle\oint (4.0\vec i -3(y^2+2)\vec j) \cdot (d\vec A)\vec j$

$\Phi= \displaystyle\oint (4.0\vec i -3(2^2+2)\vec j) \cdot (d\vec A)\vec j$

(On the top face, y=2 m because each side is 2 m in length)

$\Phi= \displaystyle\oint (4.0\vec i -18\vec j) \cdot (d\vec A)\vec j$

(Expand the brackets by taking the dot product. Remember when you take the dot product of $\vec i\cdot \vec j$ it equals 0, so our $\vec i$ term disappears)

$\Phi=-18\displaystyle\oint (d\vec A)\vec j$ $\Phi=-18A$ $\Phi=-18(2^2)$

$\Phi=-72$$\dfrac{N\cdot m^2}{C}$

b) Bottom Face

$\Phi= \displaystyle\oint (4.0\vec i -3(y^2+2)\vec j) \cdot (d\vec A)-\vec j$

(We are looking at the bottom face, so our area vector faces down, giving us negative $-\vec j$)

$\Phi= \displaystyle\oint (4.0\vec i -3(0^2+2)\vec j) \cdot (d\vec A)-\vec j$

(On the bottom face, y=0 m)

$\Phi= \displaystyle\oint (4.0\vec i -6\vec j) \cdot (d\vec A)-\vec j$

$\Phi= 6\displaystyle\oint d\vec A$

$\Phi= 6A$ $\Phi= 6(2^2)$

$\Phi= 24$$\dfrac{N\cdot m^2}{C}$

c) Left Face

$\Phi= \displaystyle\oint (4.0\vec i -3(y^2+2)\vec j) \cdot (d\vec A)-\vec i$

(Taking the dot product eliminates the j component as $\vec i \cdot \vec j = 0$)

$\Phi= \displaystyle\oint -4.0(d\vec A)$ $\Phi=-4.0A$ $\Phi=-4.0(2^2)$

$\Phi=-16$$\dfrac{N\cdot m^2}{C}$

d) Back Face

$\Phi= \displaystyle\oint (4.0\vec i -3(y^2+2)\vec j) \cdot (d\vec A)-\vec k$

(Taking the dot product eliminates both the i and j component)

$\Phi=0$

e) The net flux is equal to all the flux through each side added together. Note that the back and the front face both produce 0 flux. In addition, the right face also produces 16$\frac{N\cdot m^2}{C}$ of flux, but is positive instead of negative as our area vector faces the positive x-axis.

Net flux = -72+24-16+16+0+0 = -48 $\dfrac{N\cdot m^2}{C}$

Top face: $\Phi=-72$$\dfrac{N\cdot m^2}{C}$

Bottom face: $\Phi= 24$$\dfrac{N\cdot m^2}{C}$

Left face: $\Phi=-16$$\dfrac{N\cdot m^2}{C}$

Back face: $\Phi=0$

Net flux: $-48$$\dfrac{N\cdot m^2}{C}$

## One thought on “An electric field given by”

• Ali hassan chaudhary

Thanks a lot! May Allah bless you and give you success