Question

QUESTION 6 A closed surface with dimensions a=b=0.4 m and c=0.6 m is located as shown in the figure. If the electric field is non-uniform and given by: E = [(5.00+ 2.00x2){ – 2.509 – 2.50] N/C, where x is in meters. Calculate the Net electric flux leaving the closed surface. Y E X = a Z х +b O a. 0.269 Nm2/C O b.2.131 Nm2/C O c. 2.669 Nm2/C O d. -0.269 Nm2/C -2.131 Nm2/C O e.

Answer 1

Q= b = 0.4m 1 8. # C= 0.6m È'= (5+ 2x2){ – 2-5 ġ - 2.5€ NIC Electric flux de = dot product of electric field rector and Area Vector PE for a closed surface a uniform öğ eletric field of constant magnitude will gi've zero net electric flux. In the expression of electric field yard z components of Elletric field are constant so, Cleatic flux due to y anel z component will be zero. Only x-component op ellstic field will result in a net ellepic flux. PE = -f(x=a) Ax + Eca= gtc An PE = -(5+2(0-4)) (ab) + (5+2(0-440-603) cab) PE -(5+2(0.4)?) (0-4 x0.4)+(5+2(112) (0.480-4) PE= +0.269 N-m2/C) (9)