The cube has edge length 1.40 m and is oriented as shown in a region of uniform electric field. Find the electric flux through the right face if the electric field, in newtons per coulomb, is given by (a)6.00\vec i (b)-2.00\vec j and (c)-3.00\vec i+4\vec k (d) What is the total flux through the cube for each field?

#### Solution:

Note that when taking the dot product between unit vectors, the following rules hold:

\vec i\cdot \vec j=\vec i\cdot \vec k=\vec j\cdot \vec k=0and

\vec i\cdot \vec i=\vec j\cdot \vec j=\vec k\cdot \vec k=1

a) \Phi=(6.00)\vec i\cdot (1.4^2)\vec j=0

(0 because i⋅j=0, thus our components gets cancelled out)

b) \Phi=(-2.00)\vec j\cdot (1.4^2)\vec j=-3.92\dfrac{N\cdot m^2}{C}

c) \Phi=(-3.00\vec i+4\vec k)\cdot (1.4^2)\vec j=0

(Again, 0 because i⋅j=j⋅k=0)

d)The flux is 0. Why? Remember that **in a uniform electric field, the flux through a closed surface is 0.**

###### This question can be found in Fundamentals of Physics, 10th edition, chapter 23, question 3.