help 02 points 1 Previous Answers SePBE9 24P D19 My Notes As

help

02 points 1 Previous Answers SePBE9 24P D19 My Notes Ask Your Tea A particle with charge Q 4.80 C is located at the center of a cube of edge L 0.160 m. In addition, six other identical charged particles having q =-1.5 C are positioned symmetrically around Q as shown in the figure below. Determine the electric flux through one face of the cube. 2\'e-7 Your response differs significantly from the correct answer. Rework your solution from the baginning and check each step carefully. kN m2/c 40 L. Need Help? Resd

Solution

1.

the electric flux through one face of the cube is,

      Flux = {Q - |6q|} /6e0 = {[4.8x10^-6] - 6*(1.5x10^-6)} /(6)( 8.85x10^-12) = -7.9e+4 = -79 kN.m^2/C

Magnitude: 79 kN.m^2/C

2.

the electric potential is,

V = 4/3*pi*(R(out)^3 - R(in)^3) = 1.33*3.14(0.25^3 - 0.2^3) = 3.194*10^-2 m^3

the charge is,

Q = rho*V = (-3.46*10^-6)(3.194*10^-2) = - 1.105*10^-7 C

the net charge is,

Q(net) = Q(proton) + Q = (-60*10^-9) - (1.105*10^-7) = -1.705*10^-7

the electric field is,

E = K*Q(net) / (R out)^2 = 9*10^9*1.705*10^-7 / 0.25^2 = 2.455*10^4 N/C

the velocity is calculated as follows:

v = (e*E*R out / Mp)^0.5

= ((1.6*10^-19*2.455*10^4*0.25) / (1.67*10^-27) )^0.5

= 7.668*10^5 m/s