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Hello everyone, I've done a series of physics problems, but I have no way of seeing if they are right or wrong, So i'm going to copy and paste them, and then show you what I did to come up with the answer, I don't expect 1 person to double check all my problems, but if somtime when one or 2 of you are bored can you make sure I did them correctly? Thanks! If this isn't allowed, i'll delete the post immediatley! I Just want to make sure I get a good grade on this.

#2. A 5.0 C charge is 10 m from a -2.0 C charge. The electrostatic force is on the positive charge is:

9.0 108 N toward the negative charge

F = k (5.0C)(-2.0C)/(10m)^2 = 9.0x10^8, toward the negative charge because opposites attract.

#4. Two small charged objects repel each other with a force F when separated by a distance d. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to d/2 the force becomes:

F = (kq^2)/(d/2)

F = [(1/4)^2*2]/d = F/8

#5. What is the magnitude of a point charge that would create an electric field of 3.20 N/C at points 3.10 m away?

E = 1/(4PIEo) (q/r^2);

E(4PIEo)(r^2) = q

(3.20)(4PI)(8.85x10^-12)(3.10)^2 = q

q = 3.42x10^-9

#6. The electric field due to a uniform distribution of charge on a spherical shell is zero:

Only inside the shell

#7. In Figure 22-24, two particles of charge -q are arranged symmetrically about the y axis; each produces an electric field at point P on that axis.

http://img293.imageshack.us/img293/5161/23221om.gif [Broken]http://img293.imageshack.us/img293/5161/23221om.gif [Broken]

(a) Are the magnitudes of the fields at P equal?

yes

(b) Is each electric field directed toward or away from the charge producing it?

toward

(c) Is the magnitude of the net electric field at P equal to the sum of the magnitudes of the two field vectors (is it equal to 2E)?

no-wouldn't it be the sum of y compoents of E1 and E2?

(d) Do the x components of those two field vectors add or cancel?

cancel

(e) Do their y components add or cancel?

add

(f) Is the direction of the net field at P that of the canceling components or the adding components?

adding components

(g) What is the direction of the net field?

toward negative y

#9. Figure 22-11 shows four situations in which charged particles are fixed in place on an axis.

http://img76.imageshack.us/img76/510/22115ph.gif [Broken]http://img76.imageshack.us/img76/510/22115ph.gif [Broken]

In which situations is there a point to the left of the particles where an electron will be in equilibrium? (Select all that apply.)

situation b- visually it was the only one that made senes to me.

#10. In Figure 22-29 the electric field lines on the left have twice the separation as those on the right.

http://img339.imageshack.us/img339/83/hrw722290cb.gif [Broken]http://img339.imageshack.us/img339/83/hrw722290cb.gif [Broken]

(a) If the magnitude of the field at A is 60 N/C, what force acts on a proton at A?

E = F/q;

F = (60 N/C)(1.6x10^-19);

F = 9.6x10^-8 N;

(b) What is the magnitude of the field at B?

If the magnitude of the field at B is exactly twice as wide, would it just be 120 N/C, because A is 60 N/C ?

#11. A particle of charge of +3.40 10-6 C is 12.0 cm distant from a second particle of charge of -2.20 10-6 C. Calculate the magnitude of the electrostatic force between the particles.

F = [K(q1)(q2)]/r^2;

F = (9E9)(3.40E-6)(2.20E-6)/(.12)^2;

F = 4.675N

#12. The diagram shows two identical positive charges Q. The electric field at point P on the perpendicular bisector of the line joining them:

http://img339.imageshack.us/img339/7912/f230211bk.jpg [Broken]http://img339.imageshack.us/img339/7912/f230211bk.jpg [Broken]

Downwards- i drew the E-field lines and it looks like everyhtiung else cancels out but the y components downward.

#13. Two protons (p1 and p2) and an eletron (e) lie on a straight line, as shown. The directions of the force of p2 on p1, the force of e on p1, and the total force on p1, respectively, are:

http://img374.imageshack.us/img374/2714/f220330hu.jpg [Broken]http://img374.imageshack.us/img374/2714/f220330hu.jpg [Broken]

to the left, to the right, to the left

#14. In Figure 21-26, particle 1 of charge +1.0 µC and particle 2 of charge -2.0 µC, are held at separation L = 13.0 cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the coordinates of particle 3?

http://img291.imageshack.us/img291/8479/21262aw.gif [Broken]http://img291.imageshack.us/img291/8479/21262aw.gif [Broken]

X = ?cm

Y = ?cm

This one i am lost on, any hints would be great!!

#15. In Figure 21-24a, particles 1 and 2 have charge 45.0 µC each and are held at separation distance d = 2.70 m.

http://img291.imageshack.us/img291/9029/hrw721247yl.gif [Broken] http://img291.imageshack.us/img291/9029/hrw721247yl.gif [Broken]

(a) What is the magnitude of the electrostatic force on particle 1 due to particle 2?

F = [(9.9E9)(45.0E-6)^2]/ (2.70m)^2

F= 2.75 N

(b) In Figure 21-24b, particle 3 of charge 45.0 µC is positioned so as to complete an equilateral triangle. What is the magnitude of the net electrostatic force on particle 1 due to particles 2 and 3?

I drew a free body diagram of particle 1, and I'm alittle lost on what the angle between the two forces should be, I know 180 = sum of the angles of a triangle. So i'm thinking it should be 60. Then do i just find the resultant Force vector and thats the answer? Thanks.

Thanks everyone! Just pick and choose! any help would be great!

#2. A 5.0 C charge is 10 m from a -2.0 C charge. The electrostatic force is on the positive charge is:

9.0 108 N toward the negative charge

F = k (5.0C)(-2.0C)/(10m)^2 = 9.0x10^8, toward the negative charge because opposites attract.

#4. Two small charged objects repel each other with a force F when separated by a distance d. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to d/2 the force becomes:

F = (kq^2)/(d/2)

F = [(1/4)^2*2]/d = F/8

#5. What is the magnitude of a point charge that would create an electric field of 3.20 N/C at points 3.10 m away?

E = 1/(4PIEo) (q/r^2);

E(4PIEo)(r^2) = q

(3.20)(4PI)(8.85x10^-12)(3.10)^2 = q

q = 3.42x10^-9

#6. The electric field due to a uniform distribution of charge on a spherical shell is zero:

Only inside the shell

#7. In Figure 22-24, two particles of charge -q are arranged symmetrically about the y axis; each produces an electric field at point P on that axis.

http://img293.imageshack.us/img293/5161/23221om.gif [Broken]http://img293.imageshack.us/img293/5161/23221om.gif [Broken]

(a) Are the magnitudes of the fields at P equal?

yes

(b) Is each electric field directed toward or away from the charge producing it?

toward

(c) Is the magnitude of the net electric field at P equal to the sum of the magnitudes of the two field vectors (is it equal to 2E)?

no-wouldn't it be the sum of y compoents of E1 and E2?

(d) Do the x components of those two field vectors add or cancel?

cancel

(e) Do their y components add or cancel?

add

(f) Is the direction of the net field at P that of the canceling components or the adding components?

adding components

(g) What is the direction of the net field?

toward negative y

#9. Figure 22-11 shows four situations in which charged particles are fixed in place on an axis.

http://img76.imageshack.us/img76/510/22115ph.gif [Broken]http://img76.imageshack.us/img76/510/22115ph.gif [Broken]

In which situations is there a point to the left of the particles where an electron will be in equilibrium? (Select all that apply.)

situation b- visually it was the only one that made senes to me.

#10. In Figure 22-29 the electric field lines on the left have twice the separation as those on the right.

http://img339.imageshack.us/img339/83/hrw722290cb.gif [Broken]http://img339.imageshack.us/img339/83/hrw722290cb.gif [Broken]

(a) If the magnitude of the field at A is 60 N/C, what force acts on a proton at A?

E = F/q;

F = (60 N/C)(1.6x10^-19);

F = 9.6x10^-8 N;

(b) What is the magnitude of the field at B?

If the magnitude of the field at B is exactly twice as wide, would it just be 120 N/C, because A is 60 N/C ?

#11. A particle of charge of +3.40 10-6 C is 12.0 cm distant from a second particle of charge of -2.20 10-6 C. Calculate the magnitude of the electrostatic force between the particles.

F = [K(q1)(q2)]/r^2;

F = (9E9)(3.40E-6)(2.20E-6)/(.12)^2;

F = 4.675N

#12. The diagram shows two identical positive charges Q. The electric field at point P on the perpendicular bisector of the line joining them:

http://img339.imageshack.us/img339/7912/f230211bk.jpg [Broken]http://img339.imageshack.us/img339/7912/f230211bk.jpg [Broken]

Downwards- i drew the E-field lines and it looks like everyhtiung else cancels out but the y components downward.

#13. Two protons (p1 and p2) and an eletron (e) lie on a straight line, as shown. The directions of the force of p2 on p1, the force of e on p1, and the total force on p1, respectively, are:

http://img374.imageshack.us/img374/2714/f220330hu.jpg [Broken]http://img374.imageshack.us/img374/2714/f220330hu.jpg [Broken]

to the left, to the right, to the left

#14. In Figure 21-26, particle 1 of charge +1.0 µC and particle 2 of charge -2.0 µC, are held at separation L = 13.0 cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the coordinates of particle 3?

http://img291.imageshack.us/img291/8479/21262aw.gif [Broken]http://img291.imageshack.us/img291/8479/21262aw.gif [Broken]

X = ?cm

Y = ?cm

This one i am lost on, any hints would be great!!

#15. In Figure 21-24a, particles 1 and 2 have charge 45.0 µC each and are held at separation distance d = 2.70 m.

http://img291.imageshack.us/img291/9029/hrw721247yl.gif [Broken] http://img291.imageshack.us/img291/9029/hrw721247yl.gif [Broken]

(a) What is the magnitude of the electrostatic force on particle 1 due to particle 2?

F = [(9.9E9)(45.0E-6)^2]/ (2.70m)^2

F= 2.75 N

(b) In Figure 21-24b, particle 3 of charge 45.0 µC is positioned so as to complete an equilateral triangle. What is the magnitude of the net electrostatic force on particle 1 due to particles 2 and 3?

I drew a free body diagram of particle 1, and I'm alittle lost on what the angle between the two forces should be, I know 180 = sum of the angles of a triangle. So i'm thinking it should be 60. Then do i just find the resultant Force vector and thats the answer? Thanks.

Thanks everyone! Just pick and choose! any help would be great!

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