What is the unit vector n on the bottom face of the box? n = What is the electri

Question

What is the unit vector n on the bottom face of the box?
n =  

What is the electric flux on the bottom face of the box?
bottom =  V·m

What is the electric flux on the top face of the box?
top =  V·m

What is the electric flux on the left face of the box?
left =  V·m

What is the electric flux on the right face of the box?
right =  V·m

What is the electric flux on the front face of the box?
front =  V·m

What is the electric flux on the back face of the box?
back =  V·m

How much charge is inside the box (amount and sign)?
Qinside =  C

Explanation / Answer

Let us suppose unit vectors along x and y axes be i and j respectively.

1.) At the bottom, electric field is E = (70i + 56j)N/C

So unit vector along the field = E/lEl, where lEl = modulus of E

Now lEl = [702 + 562] = 89.64

So, unit vector along the field = E/lEl = E/89.64 = [(70/89.64)i + (56/89.64)j)]

or, unit vector along the field = 0.78i + 0.62j

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2.) Flux is = E.A

where A = area vector whose direction will be outwards on every surface of the cube. So, on the bottom surface:

A = -dwj = -(7cm)(2cm)j = -1.4X10-3m2j

E = (70i + 56j)N/C

So   =  E.A = [(70i + 56j)N/C].[ -1.4X10-3m2j] = -78.4X10-3N-m2/C

So flux on the bottom face of the box is =  -78.4X10-3N-m2/C

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3.) on the top face of the box A = dwj = (7cm)(2cm)j = 1.4X10-3m2j

E = (125i + 100j)N/C

So =  E.A = [(125i + 100j)N/C].[ 1.4X10-3m2j] = 140X10-3N-m2/C

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4.) on the left face of the box A = -dhi = -(2cm)(4cm)i = -0.8X10-3m2i

E = (125i + 100j)N/C

So =  E.A = [(125i + 100j)N/C].[ -0.8X10-3m2i] = -100X10-3N-m2/C

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This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification or correction, I will be happy to oblige....

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