InterviewSolution
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InterviewSolution
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The electric field components in Figure are E_(x)=alpha x^(1/2), E_y=E_(z)=0 in which alpha=800 N//Cm^(1/2). Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a=0.1m. |
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Answer» Solution :a. Since the electric field has only an x component, for FACES perpendicular to x DIRECTION, the angle between E and `triangleS" is "+pi/2`. Therefore, the flux `phi=E triangleS` is separately zero for each face of the CUBE except the two shaded ones. Now the magnitude of the electric field at the left face is `E_L= ax^(1/2)=alpha a^(1/2)` (x= a at the left face). The magnitude of electric field at the right face is `E_a=alpha x^((-1)/2)= alpha (2a)^(1/2)` (x=2 a at the right face). The corresponding fluxes are`phi_(L)=E_(L). triangleS=triangleS E_(L). hatn_(L)=E_(L) triangleS cos theta=-E_(L) triangleS," since "theta=180^(@), phi_(R)=E_(R). triangleS=E_(R) triangleS cos theta=E_(R) triangleS,"" sin ce"theta=0^(@)` Net flux through the cube `=phi_(R)+phi_(L)=E_(R)a^(2)-E_(L)a^(2)=a^(2) (E_(R)-E_(L)) =alpha a^(2) [(2a)^(1/2)-a^(1/2)]` `=alpha a^(5/2) (sqrt2-1) =800 (0.1)^(5/2) (sqrt2-1)=1.05 Nm^(2) C^(-1)` b. We can use Gauss. law to find the TOTAL charge q inside the cube. We have `phi=(q)/(epsi_(0)) or q=phi epsi_(0). " Therefore ",q=1.05 xx 8.854 xx 10^(-12) C=9.27 xx 10^(-12)C`. |
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