Chapter 23, Problem 002 An electric field given by E-4.7 1 4.0 y2 + 4.8) pierces

Question

Chapter 23, Problem 002 An electric field given by E-4.7 1 4.0 y2 + 4.8) pierces the Gaussian cube of edge length 0.750 m and positioned as shown in the figure (The magnitude E is in newtons per is the electricflux through the (a) top face, (b) (b) bottom face, () left coulomb and the position x s in meters.) What the cube? bo to n face, (c) left face, and (d) bac k face,te what then etele tri flu through Gaussian surface (a) Number (b) Number c) Number (d) Number (e) Number Units Units Units Units Units

Explanation / Answer

a) top face :

Area vector, A = (0.75 x 0.75) j = 0.5625 m^2 j

E = 4.7i - 4(y^2 + 4.8)j

for top face y = 0.75m and its is constant.

E = 4.7i - 4 (0.75^2 + 4.8)j = 4.7i - 21.45 j N/C

flux = E.A = - 12.06 Wb

b) bottom face

Area vector, A = (0.75 x 0.75) (-j) = - 0.5625 m^2 j

E = 4.7i - 4(y^2 + 4.8)j

for top face y = 0 and its is constant.

E = 4.7i - 4 (0+ 4.8)j = 4.7i - 19.2 j N/C

flux = E.A = 10.8 Wb

c) left face

flux due to j vector will be zero. (as field and area vector have 90 deg angle)

flux = B.A = |B||A| cos@

and cos90 = 0

due to i vector of E.

flux = 0 + ( - 0.75^2 x 4.7) = - 2.64 Wb

d) back face:

area vector = 0,75^2 (-k)

E = Exi + Ex j

so E.A = 0

flux = 0

e) flux through back and front face will be zero.

left and right face have same magnitude flux but one is -ve and other is +ve

so net flux = -12.06 + 10.8 + 2.64 - 2.64 + 0 + 0 = -1.26 Wb

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014)

Qc = 2.12 uC

Qa = - 6.37 uC

from Gauss law,

flux = Qin / e0

7.2 x 10^5 = (Qc + Qa + Qb) / (8.854 x 10^-12)

Qc + Qa + Qb = 6.37 uC

Qc - 6.37uC + 2.12uC = 6.37 uC

Qc = 10.62 uC

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