Newton's law of gravity and Coulomb's law are both inverse-
square laws. Consequently, there should be a "Gauss's law for
gravity."
a. The electric field was defined as
to find the electric field of a point charge. Using analogous
reasoning, what is the gravitational field
Write your answer using the unit vector
signs; the gravitational force between two "like masses" is
attractive, not repulsive.
b. What is Gauss's law for gravity, the gravitational equivalent
of Equation 24.18? Use
gravitational field, and Min for the enclosed mass.
c. A spherical planet is discovered with mass M, radius R, and
a mass density that varies with radius as
where
of M and R.
Hint: Divide the planet into infinitesimal shells of thickness dr,
then sum (i.e., integrate) their masses.
d. Find an expression for the gravitational field strength inside
the planet at distance r < R.
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Chapter 24 Solutions
Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
- What range of values for the introduced variable covers the rod? Replace this text with your answer. How much charge dQ is carried by a segment at location of size d? Replace this text with your answer. What is the (vector) location of dQ in terms of ? Replace this text with your answer.arrow_forwardConsider the symmetrically arranged charges in the figure, in which qa = 9 = -3.75 µC and 9c = 9a = +3.75 µC. Determine the direction of the electric field at the location of charge q. b. O right O down and right up left O up and right down and left up and left down Calculate the magnitude of the electric field E at the location of q given that the square is 6.25 cm on a side.arrow_forwardConsider two charge Q₁ = 1 x 10-6 [C] and Q₂ = 5 x 10-6 [C] separated by a distance r = 0.05 [m]. J What is the magnitude of the electric force between two charges? [N] a. F = b.₁ two charges is The electric force between the c. Suppose that the magnitude of the electric force between two charges is equal to: 27 [N]. What will be the magnitude of this force if the distance between the two charges is doubled while the charges are kept the same. F = [N] Repulsive Attractivearrow_forward
- (a) Linearize the graph. If necessary, compute different powers of variables and plot until you get a straight line. (b) Determine the equation of the line obtained. Indicate the value of n, k, and other constants or intercepts present in the graph. a.) The data below shows how the electric field (E) due to a point charge varies with distance (r). Distance, r (m) Electric Field, È (N/C) 102 25 1.25 2.50 3.75 | 5.00 | 6.25 7.50 8.75 15 9 3 10 11.25 1.15 1.08 1.005 0.6 0:.) The following values represent a particle with an x-coordinate that varies in time. Time, t (s) Distance, x (m) 3 65 1 4 5 6 7 200 195 160 -120 -425 -880 | -1515arrow_forwardConsider a particle of charge q = 2.4 C and mass m = 1.5 kg passing through the region between a pair of infinitely long horizontal plates separated by a distance d = 4.5 m with a uniform electric field strength E = 36 N/C directed in the downwards direction (-y direction). The particle begins moving horizontally with an initial velocity of v = 25 m/s from a position halfway between the plates. A.How far horizontally in meters will the particle travel before striking one of the plates. B.Caculate the speed in meters per second,with which the particle will strike the plate. C.Suppose that the eletric field is directed upward instead of downward.Caculate the new horizontal distance, in meters, that the particle travels before striking one the plates.arrow_forwardRed blood cells can often be charged. Consider two red blood cells with the following charges: -17.6 pc and +49.8 pC. The red blood cells are 3.31 cm apart. (1 pC = 1 × 10-12 C.) ..(a)..What.is.the magnitude of the force on each red blood cell? Enter a number. alculate the force of a charged particle on another charged particle? N Are the red blood cells attracted or repulsed by each other? attracted repulsed (b) The red blood cells come into contact with each other and then are separated by 3.31 cm. What magnitude of force does each of the red blood cells now experience? What is the net charge of the system? Assuming the cells are identical, how much charge will each have after contact? N Are the red blood cells attracted or repulsed by each other? attracted repulsedarrow_forward
- Electric field lines are computed with the following algorithm: 1. Pick a starting point (x.y) in space. Compute the electric field vector direction there. 2. Take a small step along this vector to another point. Compute the electric field vector at this point. Connect the previous vector to this one. 3. Repeat 2. until you step off the page, step into a charge, or reach a location where the electric field is zero. 4. Repeat 1-3 until you have enough field lines to determine the field direction in a given region of the plot. 1. A point clange y is an («,y) ** (-1,0) aKi a pa ini eharge +q is at (x,y) = (+1,0) meters. Draw an electric field line for each of the following starting points. x,y)=(0,0.1)arrow_forwardElectric field lines are computed with the following algorithm: 1. Pick a starting point (x.y) in space. Compute the electric field vector direction there. 2. Take a small step along this vector to another point. Compute the electric field vector at this point. Connect the previous vector to this one. 3. Repeat 2. until you step off the page, step into a charge, or reach a location where the electric field is zero. 4. Repeat 1-3 until you have enough field lines to determine the field direction in a given region of the plot. 1. A point charge +q is at (x.y) = (-1,0) and a point charge +q is at (x,y) = (+1,0) meters. Draw an electric field line for each of the followir.g starting points. ("SUNNY (x,y) = (-0.9,0) SUNNYarrow_forwardVery lost here. Not sure where to even begin. A small sphere with mass mm and charge qq is attached to one end of a string with length LL and tension FTFT. The other end of the string is attached to a wall which has surface-charge density σσ. The angle between the string and the wall is θθ, as shown in the drawing. a.) Please use the interactive area below to draw the Free Body Diagram for the small sphere. (I know the diagram isn't available this way but maybe a drawing?) b.) Enter an expression for the sum of the horizontal components of the forces in the diagram in terms of the variables given in the problem statement and standard physical constants. c.) Enter an expression for the sum of the vertical components of the forces in the diagram in terms of the variables given in the problem statement and standard physical constants. d.) Using the same values, L=L=9 cm, q=q=89 nC, σ=σ=3.4μC/m2μC/m2 and m=m=3.8 g, enter a numeric value for the tension in the string, FTFT, in…arrow_forward
- Q3. We have two cylindrical conductors, the inner of radius a, and outer of radius b, each infinite in extent. The outer surface of the inner conductor has a surface charge density of Ps. Use Cylindrical Coordinates to calculate D between the two cylindrical conductors, and outside the outer conductor.arrow_forwardProblem 1: What is the force F on the 1.0 nC charge at the bottom in Fig.1. Give your answer in component form. 2.OnC a) In Fig. 1, draw the vectors that represent the forces F₁, F2, and F3, exerted on the bottom charge by charges 1, 2, and 3, respectively. Pay attention to represent correctly the ratios between the magnitudes y of the forces (lengths of the vectors). Without doing calculations, try to predict the direction of the total force on the bottom charge (the total force is the vector sum of the individual forces, F = F₁+F₂ +F3). 5.0 cm 7x -6.0nC 45° 1.OnC -2.0nC 30 5.0 cm FIG. 1: The scheme for Problem 1arrow_forwardConsider a parallel plate capacitor whose plates have a charge density n. An anti-proton (a particle with the same mass as a proton, but opposite charge) starting at the positive plate is shot at a velocity of vo= 30, 000 m/s towards the negative plate. 1. What is the minimum electric field strength needed between the plates if the proton is not to hit the top plate? 2. Describe the anti-proton's trajectory. 12° Vo 45° 2.0 cm ++++arrow_forward
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