Trying to rework these problems. Thanks a lot ! A charge +4 Mu C is fixed at the

Question

Trying to rework these problems. Thanks a lot !

A charge +4 Mu C is fixed at the origin, while a charge -1 Mu C is on the y-axis at +3 m as shown. What is the resultant electric potential V at the point P on the x axis 4 m from the origin (taking V = 0 at infinity)? What is the electric field at point P (express your answer in vector notation)? If you move an electron from infinity to point P, how much work have you done? Express your answer in both eV and Joules. Be careful about signs and explain your logic carefully. A closed Gaussian surface in the shape of a cube of edge length 2 m is centered on the origin of the coordinate system (the orientation of the cube is such that the x, y, and z-axis are normal to cube faces). What is the net electric flux through the Gaussian surface?

Explanation / Answer

let q1 = +4 micro C

q2 = -2 micro C

a) Net potentail at point P,

Vnet = V1 + V2

= k*q1/d1 + k*q2/d2

= 9*10^9*4*10^-6/4 + 9*10^9*(-1*10^-6)/sqrt(4^2+3^2)

= 7200 volts

b) E1 = k*q1/d1^2 = 9*10^9*4*10^-6/4^2 = 2250 N/c

E1x = 2250 N/c

E1y = 0

E2 = k*q2/d2^2 = 9*10^9*1*10^-6/(3^2 + 4^2) = 360 N/c

E2x = -E2*cos(theta) = -360*4/5 = -288 N/c

E2y = E2*sin(theta) = 360*3/5 = 216 N/c

Enet = (E1x+E2x)i + (E1y+E2y) j

= (2250 - 288)i + (0 + 216) j

= (1962 i + 216 j) N/c

c) Potentail at infinite distance = 0

so,Workdone by exetrnal force on moving the charge = q*delta_V

= -1.6*10^-19*(7200 - 0)

= -1.152*10^-15 J

here electric field does positive work, and the workdone by us negative.

d)
According to Gauss's law,
net electric flux through the cube = Qin/epsilon

= 4*10^-6/(8.854*10^-12)

= 4.52*10^5 N/c

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