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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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

Your solution:

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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. **

STUDENT COMMENT:

Faraday explain that it reached out from the charge, so would that be a concentration? It seems to me that the concentration would be near the center of the charge and the field around it would be more like radiation extending outward weakening with distance.

INSTRUCTOR RESPONSE

That's a good, and very important, intuitive conception of nature of the electric field around a point charge.

However the meaning of the field is the force per unit charge. If you know the magnitude and direction of the field and the charge, you can find the magnitude and direction of the force on that charge.

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Question: `qExplain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane due to a given point charge at the origin.

Your solution:

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Given Solution:

`a** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin.

The electric field is therefore F / Q = k q1 / r^2. The direction is the direction of the force experienced by a positive test charge.

The electric field is therefore directly away from the origin (if q1 is positive) or directly toward the origin (if q1 is negative).

The direction of the electric field is in the direction of the displacement vector from the origin to the point if q1 is positive, and opposite to this direction if q1 is negative.

To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If q1 is positive then this is the direction of the field. If q1 is negative then the direction of the field is opposite this direction, 180 degrees more or less than the calculated angle. **

STUDENT QUESTION

Why is it just Q and not Q2?

INSTRUCTOR RESPONSE

q1 is a charge that's actually present. Q is a 'test charge' that really isn't there.  We calculate the effect q1 has on this point by calculating what the force would be if a charge Q was placed at the point in question.

This situation can and will be expanded to a number of actual charges, e.g., q1, q2, ..., qn, at specific points. If we want to find the field at some point, we imagine a 'test charge' Q at that point and figure out the force exerted on it by all the actual charges q1, q2, ..., qn.

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Question: `qQuery Principles of Physics and General Physics problem 16.15 charges 6 microC on diagonal corners, -6 microC on other diagonal corners of 1 m square; force on each.

What is the magnitude and direction of the force on the positive charge at the lower left-hand corner?

Your solution:

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Given Solution:

`a** The charges which lie 1 meter apart are unlike and therefore exert attractive forces; these forces are each .324 Newtons. This is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1 m)^2 = 324 * 10^-3 N = .324 N.

Charges across a diagonal are like and separated by `sqrt(2) meters = 1.414 meters, approx, and exert repulsive forces of .162 Newtons. This repulsive force is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6* 10^-6 C) / ( 1.414 m)^2 = 162 * 10^-3 N = .162 N.

The charge at the lower left-hand corner therefore experiences a force of .324 Newtons to the right, a force of .324 Newtons straight upward and a force of .162 Newtons at 45 deg down and to the left (at angle 225 deg with respect to the standard positive x axis, which we take as directed toward the right).

This latter force has components Fy = .162 N sin(225 deg) = -.115 N, approx, and Fx = .162 N cos(225 deg) = -.115 N.

The total force in the x direction is therefore -.115 N + .324 N = .21 N, approx; the total force in the y direction is -.115 N + .324 N = .21 N, approx.

Thus the net force has magnitude `sqrt( (.21 N)^2 + (.21 N)^2) = .29 N at an angle of tan^-1( .21 N / .21 N) = tan^-1(1) = 45 deg.

The magnitude and direction of the force on the negative charge at the lower right-hand corner is obtained by a similar analysis, which would show that this charge experiences forces of .324 N to the left, .324 N straight up, and .162 N down and to the right. The net force is found by standard vector methods to be about .29 N up and to the left. **

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Question: `qquery university physics 21.66 / 21.72 11th edition 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?

Your solution:

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Given Solution:

`a** The 6 nC charge lies 3 cm above the -2 nC charge, so it exerts force 9 * 10^9 N m^2 / C^2 * (-2 * 10^-9 C) ( 6 * 10^-9 C) / (.03 m)^2 = .00012 N.  The force between the two charges is a force of attraction, so the direction of the force on the 6 nC charge is the negative y direction.  The vector force is thus -.00012 N * `j.

The 5 nC charge lies at distance 4 cm from the 6 nC charge, so it exerts force 9 * 10^9 N m^2 / C^2 * (6 * 10^-9 C) ( 5 * 10^-9 C) / (.05 m)^2 = .000108 N, approx...  The charges repel, so this force acts in the direction of the vector 4 `i + 3 `j representing the displacement from the origin to the point (4 cm, 3 cm).  The unit vector in this direction is easily seen to be 4/5 `i + 3/5 `j = .8 `i + .6 `j.

It follows that the force vector is .000108 N ( 4/5 `i + 3/5 `j) = .000086 N * `i + .000065 N * `j.

The resultant force is therefore the sum of these two vectors, which is about .000086 N * `i - .000055 N * `j. 

This vector has magnitude sqrt( (.000086 N)^2 + (-.000055 N)^2 ) = .00011 N, approx., and angle arcTan(-.000055 N / (.000086 N) ) = -33 degrees, approx., or 360 deg - 33 deg = 327 deg.

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Question: `qQuery univ phy 21.78 / 21.80 11th edition 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)?

Your solution:

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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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Question: `qquery univ 21.104 / 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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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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