# #8a. The information below and the next 4 problems apply to one complete problem which is broken up into parts as discussed in the class.?

A thin straight glass rod, 550 cm long, is placed on the x-axis so that it is symmetric about the y axis. An electrical charge Q = -328 mC, is uniformly distributed over it. You will determine the electric field at a point P located at (0.00, 6.50 m). The following takes you step by step working through the problem. This gives practice in using the steps to solve this type of problem.

The charge density of the rod is

Answer:_________

. Give your answer in the form "+/-a.bc x 10^(y) C/m". (Do you agree that the SI unit is C/m? Be sure that you understand that this unit is correct!)

8b. As was done in class, choose a differential segment of length dl on left section of the rod. Call this tiny segment dx since it is on the x-axis. Remember that this is a length. Also, let this dx be an arbitrary distance x from the origin.

The differential amount of charqe dQ on dx is expressed as dQ =

Answer :________

dx . Fill in the blank. Give your answer in the form "+/-a.bc x 10^(d) unit". Hint. remember that dQ is electric charge so the unit must be the unit for charge.

8c. The differential amount of charqe dQ on dl will setup a tiny electric field dE at point P . Sketch this field at P before going any further. Since there is symmetry in the problem (do you agree with this?) then it should be possible to find an identical dQ, the same distance from the y-axis as the original dQ that you had chosen on the other side of the rod. This will also set up a tiny electric field dE at P. Sketch this field also at P. Which of the following would be true?

Note that there are more than one correct answers. Give them all. Give your answer in the form "M", "N" where N and M represent the answer choices. Note that a comma followed by a space is used to separate the letters.

A. The y components of the two dE's will cancel out leaving only the x components.

B. The x components of the two dE's will cancel out leaving only the y components.

C. Because of the symmetry the components that do not cancel out to give zero will be identical.

D. Because of the symmetry the components that do not cancel out to give zero will be not be identical.

8d. Because of the symmetry you now only need to integrate the surviving component of dE for the dQ's on only one half of the rod and then multiply your result by 2 in order to get the magnitude of the electric field E at point P due to the Q charge on the rod. (You should be convinced that that is true.)

Complete the integration to complete the calculation of the magnitude of the electric field E at point P. Give your answer in the form "a.bc x 10^(n)" N/C

8e. Complete the problem now by giving the magnitude and direction of E at P. Give your answer in the form "a.bc x 10^(x) N/C", in the "+/- X or Y" axis direction. For example if the magnitude of the field is 3.95 x 10^(5) N/C and the direction is in the -x axis direction then your answer input should be

3.95 x 10^(5) N/C, -X

### 1 Answer

- WoodsmanLv 72 months agoFavorite Answer
8a. λ = Q/L = -0.328C / 0.550m = -0.596 C/m

so your format is -5.96 x 10^(-1) C/m

8b. dQ = λ*dx = -5.96 x 10^(-1) C

per meter of rod

8c. See citation for an image. The charge is negative, so the field lines point from P to the element dL.

A - false

B - true

C - true

D - false

so in your format,

B, C

8d. dE = k*λ*z*dx / (z²+x²)^1.5

E = k*λ*z*∫ dx / (z²+x²)^1.5

evaluated from -L/2 to L/2, or simply from 0 to L/2 and then double it:

E = 2*k*λ*z*∫ dx / (z²+x²)^1.5

from 0 to L/2.

E = (2*k*λ/z)*(L/2)/√(z²+(L/2)²)

The expression (L/2)/√(z²+(L/2)²) is dimensionless, and

2*k*λ/z has units N·m²/C² * C/m / m = N/C

and so this is looking like a sound result.

Plugging in k = 8.99e9 N·m²/C²

and |λ| = 0.596 C/m

and z = 6.50 m

and L/2 = 0.275 m

I get

E = 6.97e7 N/C

so in your format 6.97 x 10^7

Note that if we consider the rod as a point charge we get

E = kQ/z² = 8.99e9 N·m²/C² * 0.328C / (6.50m)² = 6.98e7 N/C

and so this should boost your confidence in the result of the integration

8e. You have the magnitude from part d. The direction is STRAIGHT DOWN (-Y)

since the x-components cancel and the charge is negative.

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