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The electricfield componenets in Fig are E_(x) = alpha x^(1//2), E_(y) = E_(z) = 0 inwhich alpha = 800 N//C -m^(1//2). Considerthe cubeshown in Fig. Calculate(a) the flux phi_(E)throughthe cube,and(b) the charge within the cube. Assumethat a = 0.1m. |
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Answer» Solution :(a) Here, `E_(x) = ax^(1//2), E_(y) = 0, E_(z) = 0` `alpha = 800 N//C -m^(1//2) , a = 0.1m`. As the electricfieldhas onlyx componenet, therefore.`E. Delta vec(S) = phi_(E) = 0` for each of four faces of cube`_|_` to Y-axis and Z-axis. Flux is there only for left face L and rightface Rof the cubeshown in Fig. At the left face, x = a `:. E_(L) = alpha a^(1//2)` `phi_(L) = vec(E_(L)) . Delta vec(S) =alpha a^(1//2) (a^(2)) cos 180^(@) = -alpha a^(5//2)` At the right face, `x = a + a = 2a` ':. E_(R) = alpha (2a)^(1//2)` `phi_(R) = vec(E_(R)) . Delta vec(S) = alpha (2a)^(1//2) (a^(2)) cos 0^(@)` `=alpha a^(5//2) sqrt(2)` `:.` NET flux through the cube `= phi_(R) + phi_(L)` `= alpha a^(5//2) sqrt(2) - alpha a^(5//2)` `phi = alpha a^(5//2) (sqrt(2) - 1)= 800 (0.1)^(5//2) (sqrt(2) - 1)` `= 1.05 N m^(2) C^(-1)` (b) By GAUSS's theorem, `q = in_(0) phi = 8.85 xx 10^(-12) xx 1.05` `= 9.27xx10^(-12)C` |
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