An electric field given by [ E = 6.8?i - 6.3(y^2 + 5.3)?j ] pierces a Gaussian c
ID: 2149540 • Letter: A
Question
An electric field given by [ E= 6.8?i - 6.3(y^2 + 5.3)?j ] pierces a Gaussian cube of edge length 5.2 m and positioned as shown in Fig. 23-56. (The magnitude E is in newtons per coulomb and the position x is in meters.) What net charge (in Coulombs) is enclosed by the Gaussian cube?
An electric field given by [ E= 6.8?i - 6.3(y^2 + 5.3)?j ] pierces a Gaussian cube of edge length 5.2 m and positioned as shown in Fig. 23-56. (The magnitude E is in newtons per coulomb and the position x is in meters.) What net charge (in Coulombs) is enclosed by the Gaussian cube?Explanation / Answer
We use the Gaussian law Integral of E.dA over all 6 surfaces of the cube = q/epsilon0 Consider the two surfaces with normals parallel to x axis net E = 6.8A - 6.8A =0 Now consider the surface with area l*l j E.A = -6.3(5.2^2 + 5.3)*5.2^2 (y=l=5.2) Now consider the surface with area vector -l*l j Flux = E.A = 6.3(0^2 + 5.3)*5.2^2 (y=0) SO Net flux = -6.3(5.2^2 + 5.3)*5.2^2 + 6.3(0^2 + 5.3)*5.2^2 = -4606.3 So Q = -8.85 * 10^-12 *4606.3 = -40.77 * 10^-9 Coloumb Hope you like it :)
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