why did I get the wrong answer??? k e x ( x 2 + a 2 ) 3/2 Consider a uniformly c

Question

why did I get the wrong answer???

kex (x2 + a2)3/2 Consider a uniformly charged thin-walled right circular cylindrical shell having total charge Q, radius R, and length l. Determine the electric field at a point a distance d from the right side of the cylinder as shown in the figure below. Suggestion: Use the following expression and treat the cylinder as a collection of ring charges. (Use any variable or symbol stated above along with the following as necessary: ke.) E = 2k e q/r 2 l (1 - ((d + l) 2 + r 2) 0.5 + (d 2 + r 2) 0.5) Consider now a solid cylinder with the same dimensions and carrying the same charge, uniformly distributed through its volume. Use the following expression to find the field it creates at the same point. (Use any variable or symbol stated above along with the following as necessary: ke.)

Explanation / Answer

A uniformly charged thin-walled right circular cylindrical shell has a total charge Q, radius R, and height h, and is aligned with the x-axis.

a) Determine the electric field at a point a distance d from the right side of the cylinder
b) Consider now a solid cylinder with the same dimensions and carrying the same charge, uniformly distributed through its volume. Determine the electric field at a point a distance d from the right side of the cylinder.

(like what's a uniformly charged thin-walled right circular cylindrical shell???!!! and where is the equation for that?!!) =(

..........................

The thin-walled, etc is just like a can with the ends removed. The charge is uniformly distributed across the surface. The area of the cylinder is just:
Area = 2*pi*R*h

Charge per unit area = Q/Area = Q/[2*pi*R*h]

a. I am going to assume that they mean the right end of the cylinder and guess that they want the electric field at a point on the x-axis that is a distance d from the end of the cylinder. I am also going to assume that the middle of the cylinder is at x = 0. This means that the net electric field will be pointing along the x-axis only.

Find the electric field from a point on the surface at the point d.

dE = cos(A)kdQ/r^2 ..... the cosine because there are components both parallel to the x-axis and perpendicular and we only need the parallel part.

k = constant
A = angle the vector from the surface point to d makes with the x-axis
r = distance from point on the surface to the point at d which can be found using the standard distance between two points.
diff in y = R
diff in x = d + h/2 - x
r = SQRT[R^2 + (d + h/2 - x)^2]

We can use a ring on the surface because of symmetry and the charge becomes;
Area of ring = 2*pi*R*dx
dQ = [2*pi*R*dx]Q/[2*pi*R*h] = (Q/h)dx

We also need the angle A.
cos(A) = (d + h/2 - x)/SQRT[R^2 + (d + h/2 - x)^2]

E is defined as kQ/r^2 so we can write for dE:
dE = cos(A)kdQ/r^2
dE = k{(d + h/2 - x)/SQRT[R^2 + (d + h/2 - x)^2]}{(Q/h)dx}/[R^2 + (d + h/2 - x)^2]

and this is integrated from x = -h/2 to x = h/2

dE = (kQ/h)(d + h/2 - x){dx}/[R^2 + (d + h/2 - x)^2]^(3/2)

Let w = R^2 + (d + h/2 - x)^2
dw = -2(d + h/2 - x)dx

dE = [-kQ/(2h)] dw/w^(3/2)
E = [-kQ/(2h)] [-2/SQRT(w)]
E = (kQ/h)[1/SQRT[R^2 + (d + h/2 - x)^2] ... x = -h/2 to x = h/2
E = (kQ/h){[1/SQRT(R^2 + d^2)] - [1/SQRT(R^2 + (d + h)^2)]}

b. The charge per unit volume is:
V = volume of cylinder = pi^R^2h
C = charge per unit volume = Q/[pi*R^2h]

We will now need to integrate the ring from part a from 0 to R to get the effect for a slice of the cylinder and then integrate that from x = -h/2 to x = h/2.

Element of volume for ring = dV
dV = 2*pi*RdRdx
Charge = dQ = (charge per unit volume)* (volume)
dQ =C*[2*pi*RdRdx]

And similar to part a we can write for dE:
dE = cos(A)kdQ/r^2
dE = (d + h/2 - x)/SQRT[R^2 + (d + h/2 - x)^2]k{C*[2*pi*RdRdx]} /[R^2 + (d + h/2 - x)^2]
dE = [d + h/2 - x]{kC*[2*pi*RdRdx]} /[R^2 + (d + h/2 - x)^2]^(3/2)

First integrate over R and evaluate from R = 0 to R.
dE = [kC*pi (d + h/2 - x)dx] {2RdR / [R^2 + (d + h/2 - x)^2]^(3/2)}

w = R^2 + (d + h/2 - x)^2
dw = 2RdR
Integral of dw/w^(3/2) = -2/SQRT(w) = -2/SQRT[R^2 + (d + h/2 - x)^2]
Integral = (-2){1/SQRT[R^2 + (d + h/2 - x)^2] - 1/(d + h/2 - x)}

Now integrate over x and evaluate from x = -h/2 to x = h/2.
[kC*pi (d + h/2 - x)dx] (-2){1/SQRT[R^2 + (d + h/2 - x)^2] - 1/(d + h/2 - x)}

The first term is:
[-2kC*pi (d + h/2 - x)dx]/SQRT[R^2 + (d + h/2 - x)^2]

Let w = R^2 + (d + h/2 - x)^2
dw = -2(d + h/2 - x)dx

(kC*pi) dw/SQRT(w)

Integrate to get: (2kC*pi)SQRT(w) = (2kC*pi)SQRT[R^2 + (d + h/2 - x)^2]
Evaluate: (2kC*pi){SQRT[R^2 + d^2] - SQRT[R^2 + (d + h)^2]}

The second term is:
[kC*pi (d + h/2 - x)dx] (2)/(d + h/2 - x) = 2kC*pi*dx
Integrated this is just 2kC*pi[x]
Evaluated: (2kC*pi)[h/2 - (-h/2)] = 2kC*pi*h

Put the terms together:
E = 2kC*pi*h +(2kC*pi){SQRT[R^2 + d^2] - SQRT[R^2 + (d + h)^2]}
E = 2kC*pi*{h + SQRT[R^2 + d^2] - SQRT[R^2 + (d + h)^2]}

Now put in the expression for C:
E = 2k[Q/(pi*R^2h)]*pi*{h + SQRT[R^2 + d^2] - SQRT[R^2 + (d + h)^2]}
E = 2k[Q/(R^2h)]{h + SQRT[R^2 + d^2] - SQRT[R^2 + (d + h)^2]}

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.