HELP!! ASAP!! Need help with Physics!!! thanks :)?
1) Suppose a charge q is placed at point x = 0, y = 0. A second charge q is placed at point x = 7.4 m, y = 0. What charge Q must be placed at the point x = 3.7 m, y = 0 in order that the field at the point x = 3.7 m, y = 4.5 m be zero?
Q = answer in q
2) An electron (q = -1.602 10-19 C) is projected horizontally into the space between two oppositely charged metal plates. The electric field between the plates is 547.5 N/C, directed up.
(a) While in the field, what is the force on the electron?
magnitude: answer in N
(b) If the vertical deflection of the electron as it leaves the plates is 3.80 mm, how much has its kinetic energy increased due to the electric field?
answer in J
3) A conducting sphere is placed within a conducting spherical shell as shown in the figure below. The conductors are in electrostatic equilibrium. The inner sphere has a radius of 1.50 cm, the inner radius of the spherical shell is 2.25 cm, and the outer radius of the shell is 2.75 cm. The inner sphere has a charge of 275 nC, and the spherical shell has zero net charge.
(a) What is the magnitude of the electric field at a point 1.75 cm from the center?
answer in N/C
(b) What is the electric field magnitude at a point 2.50 cm from the center? [Hint: What must be true about the electric field inside a conductor in electrostatic equilibrium?]
answer in N/C
(c) What is the electric field magnitude at a point 3.00 cm from the center?
answer in N/C
please help me out with these questions..i really need help to understand this questions in order for me to understand physics for my oncoming exam...please show steps...best answers will get 10 pts..thank you very much greatly aprreciated
- DrostieLv 610 years agoFavorite Answer
0. We would really rather you ask questions about material that you don't understand, rather than asking us to do your homework for you.
1. There is a nice reflection symmetry about the x = 3.7 m line, so you only have to calculate the y-component of the force. The distance from either point to (3.7, 4.5) is given by the Pythagorean theorem as r = 5.8 m. Given this projection onto the y-axis, we will want to multiply by the cosine of the angle we're creating, and this cosine is adjacent/hypotenuse = y / r = 4.5 m / 5.8 m.
The coulomb field is k q / r², where k is a force constant depending on your unit system, q is the magnitude of the charge, and r is the distance from the charge q.
Combining the two charges, the total y-component of the E-field at the point will therefore be:
2 * k q/r² * y/r = 2 k q y / r³.
The charge that we're adding, Q, must balance out this with its own force:
2 k q y / r³ + k Q / y² = 0
Q = - 2 q (y / r)³
2. a. The Lorentz force is F = q E. You're given q and E. Multiply them.
b. work is W = F Δr. You just got F and now you're given Δr. Multiply them.
3. a. You may ignore the effect of the outer conductor within this space, and the mathematics allows a spherically symmetric charge distribution to be treated as an equivalent point charge at its center. So, use your k q / r² formula here, with r = 1.75 cm.
b. A good conductor will have the same voltage at every point, and therefore no E-field (since E is the gradient of V).
c. Again, collapse all of the spherically symmetric charge distributions to point charges at their center, and use k q / r².
- 10 years ago
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