Problems that I got wrong and still don't understand

[jst]

Before my discovery of this forum, I tried to do some homwork with not help...well, I got a few wrong and will be having a quiz soon...so I got to correct a few of my old mistakes

Homework Statement

Given a uniformly charged arc with radius R and total charge Q. The arc subtends a total angle of 90 degrees. Derive an equation for the total electric field at the center of curvature of this arc in terms of R,Q and the Coulomb constant k.

http://glomawr.com/phy-hw-01-02.jpg

Homework Equations

I'm not sure

The Attempt at a Solution

I put: (kQ^2)/R^2 , which was not the correct answer

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Homework Statement

Given: two charge diagrams each consisting of three charges arranged in an equilateral triangle. Each charge has the same magnitude of charge, q. Force directions are labeled A through L spaced at 30 degree intervals. What is the direction of the Coulomb force on charge #3 in Diagram A?

http://glomawr.com/phy-hw-01-04a.jpg

Homework Equations

The Attempt at a Solution

I put "B" and was correct, but I'm not sure my logic was correct.

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Homework Statement

Given two charges aligned horizontally. Initially both charges have an equal charge -q. The electric force on the right charge is shown having a magnitude of one. In the diagrams below changes have been made to the situation. For each diagram A,B, and C, state the relative length of the electric force vector on the right charge.

http://glomawr.com/phy-hw-01-08.jpg

Homework Equations

The Attempt at a Solution

I put: A half, B 4, C 4

The correct was: A 8, B 4, C half
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Homework Statement

Given: Two point charges of charge +q and - q aligned horizontally. The positive charge feels an electrostatic force F. Five points along the horizontal axis are labeled as shown A through E. To what point could a third charge +q be placed so that the net electrostatic force on the positive charge is 3F?

http://glomawr.com/phy-hw-01-09.jpg

Homework Equations

The Attempt at a Solution

I put A, but the correct answer was B

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Homework Statement

Given: Two point charges of charge 7.00 nC and -7.00 nC aligned horizontally. The charges are separated by a distance of 8.00 cm. What is the electric field strength at point B halfway between the charges?

http://glomawr.com/phy-hw-01-10.jpg

Homework Equations

The Attempt at a Solution

I did:

(8.988*10^9)*(7*10^-9)/(.08/2) +
(8.988*10^9)*(-7*10^-9)/(.08/2)
=0

Which was wrong, the solution was:
.0788 MN/C

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Homework Statement

Given a uniformly charged vertical line and a uniformly charged thin half ring with their axes of symmetry aligned horizontally as shown. The total charge on the line is Q. The total length of the line is 2.00 m and the half-ring radius is R = 1.00 m. At a distance x = 1.00 m to the right of the line charge is the center of curvature of the half ring. What is the total charge on the half ring if the total electric field at the center of curvature is zero? (You will need to integrate over the half ring. S = R(theta) and dS = R(dtheta) and dq = (charge per length)(dS)

http://glomawr.com/phy-hw-01-11.jpg

Homework Equations

The Attempt at a Solution

I some how came up with: 3.84Q

But the solution was: 1.11Q


I appreciate any help with any/all of these.

Thanks,

Jason

 

[member 553083]

For the first question, the equation for the total electric field at the center of curvature is E= (kQ)/(2R).For the third question, the correct answer is B because Coulomb's Law states that the force between two charges is in the direction from positive to negative, so the force on charge #3 in Diagram A is in the direction "B".For the fourth question, the correct answer is A: 8, B: 4, and C: 1/2, as the electric force vector will be 8 times as long on A than C. This is because the force between two charges is inversely proportional to the square of their distances, so the electric force vector on A is 8 times as long as the electric force vector on C.For the fifth question, the correct answer is B because the net electrostatic force on the positive charge is 3F when the third charge +q is placed at point B. This is because the net force on the positive charge is the vector sum of the two individual forces, so when the third charge +q is placed at point B, it will equal and cancel out the force from the negative charge, resulting in a net force of 3F on the positive charge.For the sixth question, the equation for the electric field strength at point B is E = (kQ)/(4x^2). Plugging in the given values, you get E = (8.988*10^9)(7*10^-9)/(4*1)^2 = 0.0788 MN/C.