Assignment 24

course Phy 232

July 15 at 11:15

024.

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Question: `qIn your own words explain the meaning of the electric field.

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Your solution:

An electric field is the amount of force per charge in the space around a charged particle. It can push or pull particle away from each other. The higher the force due to two charged particles, the higher the electric field will be.

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Given Solution:

`aSTUDENT RESPONSE AND INSTRUCTOR COMMENT: electric field is the concentration of the electrical force

** That's a pretty good way to put it. Very specifically electric field measures the magnitude and direction of the electrical force, per unit of charge, that would be experienced by a charge placed at a given point. **

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Self-critique (if necessary):

Self-critique Rating:

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Question: `qquery university physics 21.68 (22.52 10th edition) 5 nC at the origin, -2 nC at (4 cm, 0).

If 6 nC are placed at (4cm, 3cm), what are the components of the resulting force?

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Your solution:

F_x = F_x1 + F_x2, F_x2 = 0

F_x = F_x1 =( 9*10^9)*(5*10^-6 C)(6*10^-6 C)*(.04 m)/(0.05 m)^3 = 86.4 N

F_y = F_y1 + F_y2

F_y = 9*10^9(6*10^-6 C)*(5*10^-6 C * 0.03 m/(0.05 m)^3 – (2*10^-6 C)/(0.03^2))

F_y = -55.2 N

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Given Solution:

`a** The 5 nC charge lies at distance sqrt( 4^2 + 3^2) cm from the 6 nC charge, which is 3 cm from the charge at (4 cm ,0).

The force exerted on the 6 nC charge by the two respective forces have magnitudes .00011 N and .00012 N respectively.

The x and y components of the force exerted by the charge at the origin are in the proportions 4/5 and 3/5 to the total charge, giving respective components .000086 N and .000065 N.

The force exerted by the charge at (4 cm, 0) is in the negative y direction.

So the y component of the resultant are .000065 N - .00012 N = -.000055 N and its x component is .000086 N. **

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Self-critique (if necessary):

My process was right, my answer is off by a magnitude because I used the wrong unit conversion for nC.

Self-critique Rating:

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Question: `qQuery univ phy 21.76 (10th edition 22.60) quadrupole (q at (0,a), (0, -a), -2q at origin).

For y > a what is the magnitude and direction of the electric field at (0, y)?

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Your solution:

E = F/q = (F1+F2+F3)/q_y

F1 = k*q*q_y/(y-a)^2

F2 = -2*k*q*q_y/y^2

F3 = k*q*q_y/(y+a)^2

Plugging these into the first equation gives us:

E = k*(q/(y-a)^2 – 2*q/y^2+q/(y+a)^2)

This simplifies to:

E = 2*k*q/a^2 #######

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Given Solution:

`a** The magnitude of the field due to the charge at a point is k q / r^2.

For a point at coordinate y on the y axis, for y > a, we have distances r = y-a, y+a and y.

The charges at these distances are respectively q, q and -2q.

So the field is

k*q/(y - a)^2 + k*q/(y + a)^2 - 2k*q/y^2 = 2*k*q*(y^2 + a^2)/((y + a)^2*(y - a)^2) - 2*k*q/y^2

= 2*k*q* [(y^2 + a^2)* y^2 - (y+a)^2 ( y-a)^2) ] / ( y^2 (y + a)^2*(y - a)^2)

= 2*k*q* [y^4 + a^2 y^2 - (y^2 - a^2)^2 ] / ( y^2 (y + a)^2*(y - a)^2)

= 2*k*q* [y^4 + a^2 y^2 - y^4 + 2 y^2 a^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [ 3 a^2 y^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) .

For large y the denominator is close to y^6 and the a^4 in the numerator is insignifant compared to a^2 y^2 sothe expression becomes

6 k q a^2 / y^4,

which is inversely proportional to y^4. **

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Self-critique (if necessary):

I made a huge algebraic mistake by saying that a/(b+c) = a/b + a/c

Self-critique Rating:

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Question: `qquery univ 22.102 annulus in yz plane inner radius R1 outer R2, charge density `sigma.What is a total electric charge on the annulus?

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Your solution:

Charge Density (sigma) = q/A

A = 2(pi(R2^2 – R1^2)) I am assuming we are talking about both sides of the disk.

Q = 2*sigma*(pi(R2^2 – R1^2))

Next we want to find the magnitude and direction of the electric field at an arbitrary point on the x-axis.

How are the charge, distance, and shape of the annulus related?

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Given Solution:

`a** The total charge on the annulus is the product

Q = sigma * A = sigma * (pi R2^2 - pi R1^2).

To find the field at distance x along the x axis, due to the charge in the annulus, we first find the field due to a thin ring of charge:

The charge in a thin ring of radius r and ring thickness `dr is the product

`dQ = 2 pi r `dr * sigma

of ring area and area density.

From any small segment of this ring the electric field at a point of the x axis would be directed at angle arctan( r / x ) with respect to the x axis. By either formal trigonometry or similar triangles we see that the component parallel to the x axis will be in the proportion x / sqrt(x^2 + r^2) to the magnitude of the field from this small segment.

By symmetry only the xcomponent of the field will remain when we sum over the entire ring.

So the field due to the ring will be in the same proportion to the expression k `dQ / (x^2 + r^2).

Thus the field due to this thin ring will be

magnitude of field due to thin ring: k `dQ / (x^2 + r^2) * x / sqrt (x^2 + r^2) = 2 pi k r `dr * x / (x^2 + r^2)^(3/2).

Summing over all such thin rings, which run from r = R1 to r = R2, we obtain the integral

magnitude of field = integral ( 2 pi k r x /(x^2 + r^2)^(3/2) with respect to r, from R1 to R2).

Evaluating the integral we find that

magnitude of field = 2* pi k *x* | 1 /sqrt(x^2 + r1^2) - 1 / sqrt(x^2 + r2^2) |

The direction of the field is along the x axis.

If the point is close to the origin then x is close to 0 and x / sqrt(x^2 + r^2) is approximately equal to x / r, for any r much larger than x. This is because the derivative of x / sqrt(x^2 + r^2) with respect to x is r^2 / (x^2+r^2)^(3/2), which for x = 0 is just 1/r, having no x dependence. So at small displacement `dx from the origin the field strength will just be some constant multiple of `dx. **

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Self-critique (if necessary):

Self-critique Rating:

&#This looks good. Let me know if you have any questions. &#