I just don\'t understand part A, I need a step by step solution. CH 26 HW Proble

Question

I just don't understand part A, I need a step by step solution.

CH 26 HW Problem 26.48 Problem 26.48 Consider a large, thin, electrically neutral conducting plate in the zy-plane at -0 and a point charge gon the z-axis at distance echarge d. What is the electric field on and above the plate? Although the plate is neutral, electric forces from the point charge polarize the conducting plate and cause it to have some complex distribution of surface charge. The electric field and potential are then a superposition of fields and potentials due to the point charge and the plate's surface charge. That's complicated! However, it is shown in more advanced classes that the field and potential outside the plate (z 0) are exactly the same as the field and potential of the original charge plus a "mirror image" charge located at Resources previous l 10 of 14 l next a You completed this assignment. Part A Select the correct expression for electric potential in the yz-plane forz20 (i e. in the space above the plate). the VO y, z Kg Submit My Answers Give Up Correct Part B We know that electric fields are perpendicular to conductors in electrostatic equilibrium, so the field at the surface of the plate has only a z-component. Select the correct expression for the field E. on the surface of the pla 00 as a function of distance y away from the 2 Kg

Explanation / Answer

Part A

As stated in the problem, you can replace the entire plate with a charge q located at z = d and a mirror charge -q at z = -d for the purpose of calculating potential in z > 0

We have to calculate potential in region z > 0, in the y-z plane, so, here x = 0

Now, let us take a point P (0,y,z) in the region z > 0 where we want to calculate the potential due to the charges q and -q at z = d and z = -d.

Position of charge q is, Q1(0,0, d),

So, PQ1 = [02 + (y-0)2 + (z - d)2]1/2 = [(d - z)2 + y2 ]1/2 ................ (as , (d-z)2 = (z-d)2)

So potential at P(0,y,z) due to q is,

V1 = (1/40)q/[(d - z)2 + y2 ]1/2 = Kq/[(d - z)2 + y2 ]1/2, where K = 1/40

Similarly, Position of charge -q is, Q2(0,0, -d), so PQ2 =  = [02 + y2 + (z + d)2]1/2

So potential at P(0,y,z) due to -q is,

V2 = (1/40)(-q)/[(d + z)2 + y2 ]1/2 = -Kq/[(d + z)2 + y2 ]1/2, so net potential at P is,

V = V1 + V2 =  Kq/[(d - z)2 + y2 ]1/2 + (-Kq/[(d + z)2 + y2 ]1/2)

or, V = Kq[1/[(d - z)2 + y2 ]1/2 - 1/[(d + z)2 + y2 ]1/2]

***************************************************************************************************
This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification, modification or correction, feel free to ask.....

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---