Solving Homework Problems: Electric Field & Gravitational Field

In summary, electric and gravitational fields are two types of forces that affect the motion of objects. Electric fields are caused by the interaction between charged particles and are represented by a vector quantity, while gravitational fields are caused by the interaction between masses and are represented by a scalar quantity. The electric field at a point is calculated by dividing the force on a test charge by the magnitude of the test charge, and the formula for gravitational field is given by Gm/r^2. Both fields can cause objects to accelerate and change their motion, and to solve problems involving them, one must use the appropriate equations and principles while paying attention to units and using proper mathematical techniques.
  • #1
bananasplit
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Homework Statement


Problem 1. A block having mass m and charge +Q is connected to an
insulating spring having force constant k. The block lies on a frictionless, insulating horizontal track, and the system is immersed in a uniform electric field of magnitude E directed as shown in Figure
P25:7. The block is released from rest at a moment when the spring is unstretched (that is, when x = 0).
(a) By what maximum amount does the spring expand?
(b) What is the equilibrium position of the block?

Problem 2. Consider a closed surface S in a region of gravitational field g. Gauss’s law for gravitation tells us that the gravitational flux through surface S is linearly proportional to the total mass min occupying the volume contained by S. More specifically, Gauss’s law states that
(closed integral)g x da = -4Gmin :
Note that g here is the total electric field, due to mass sources both inside and outside S. The value of G, the gravitational constant, is about 6.673 x10-11 N m2/kg2.
(a) Earth’s volume mass density, at any distance r from its center, is given approximately by the function p = A-Br=R, where A = 1.42 x 104 kg/m3, B = 1.16 x 104 kg/m3, and Earth’s radius R = 6.370 x 106 m. Calculate the numerical value of Earth’s mass M. Hint: The volume of a
spherical shell, lying between radii r and r + dr, is dv = 4(pie)r2dr.
(b) Determine the gravitational field inside Earth.
(c) Using the result of part b, determine the gravitational field magnitude at Earth’s surface.


Homework Equations





The Attempt at a Solution


1.Arbitrarily choose V = 0 at 0. Then at other points
V= −Ex and Ue =QV=−QEx.
Between the endpoints of the motion,
(K +Us+ Ue)i = (K+ Us +Ue)f
0+0+0=0+(1/2)kx2max −QExmax so xmax = (2QE)/k

At equilibrium,
ΣFx= −Fs+Fe= 0 or kx =QE .
So the equilibrium position is at x = QE/k

Problem 2. I have no clue at where to begin, or what equations to use. Any help is appreciated.
 
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  • #2


Thank you for posting this problem. I am happy to assist you in finding the solutions for these two problems.

For Problem 1, we can use the equations for the potential energy and the equilibrium position to solve for the maximum expansion of the spring and the equilibrium position of the block. The potential energy of the system is given by U = (1/2)kx^2 + QEx, where k is the spring constant, x is the displacement of the block, Q is the charge of the block, and E is the electric field. At the maximum expansion, the potential energy is equal to the maximum potential energy, which is (1/2)kx^2max. We can equate these two and solve for x^2max. Similarly, at the equilibrium position, the potential energy is equal to zero, since the block is not displaced. We can use this to solve for the equilibrium position, xeq. The equations you have used in your attempt are correct, and the final answers should be: xmax = (2QE)/k and xeq = QE/k.

For Problem 2, we need to use Gauss's law for gravitation to solve for the total mass inside the closed surface S. This law states that the gravitational flux through a closed surface is equal to -4Gm, where G is the gravitational constant and m is the total mass inside the surface. We can use this to solve for the mass M of the Earth, by setting the gravitational flux equal to the gravitational field (g) times the surface area (da). The rest of the problem involves using the given equations and the mass M to solve for the gravitational field inside and at the surface of the Earth. The final answers should be: M = 5.98 x 10^24 kg, g = GMr/R^3 (inside Earth), and g = GM/R^2 (at Earth's surface).

I hope this helps. Let me know if you have any further questions. Keep up the good work!
 
  • #3


As a scientist, it is important to have a strong understanding of both electric and gravitational fields and how they interact with matter. In problem 1, we are given a system consisting of a block with mass and charge connected to a spring, immersed in a uniform electric field. To solve this problem, we can use the equations for electric potential and energy, as well as the principles of conservation of energy and equilibrium. By setting up the equations and solving for the maximum amount the spring expands and the equilibrium position of the block, we can find the answers to parts (a) and (b).

In problem 2, we are dealing with a closed surface in a region of gravitational field. To solve this problem, we can use Gauss's law for gravitation, which tells us that the gravitational flux through a closed surface is proportional to the total mass contained within that surface. By using the given values for Earth's volume mass density and radius, we can calculate the numerical value of Earth's mass in part (a). In part (b), we can determine the gravitational field inside Earth by using the equation for gravitational field strength. Finally, in part (c), we can use the result from part (b) to find the gravitational field magnitude at Earth's surface. By using the appropriate equations and principles, we can solve problem 2 and gain a better understanding of gravitational fields.
 

1. What is the difference between electric field and gravitational field?

The main difference between electric field and gravitational field is the type of force they represent. Electric field is caused by the interaction between charged particles, while gravitational field is caused by the interaction between masses. Additionally, electric field is a vector quantity, meaning it has both magnitude and direction, while gravitational field is a scalar quantity, only having magnitude.

2. How do you calculate the electric field at a point?

The electric field at a point is calculated by dividing the force on a test charge placed at that point by the magnitude of the test charge. This can be represented by the equation E = F/q, where E is the electric field, F is the force, and q is the test charge.

3. What is the formula for gravitational field?

The formula for gravitational field is given by Gm/r^2, where G is the universal gravitational constant, m is the mass of the object creating the field, and r is the distance from the object to the point where the field is being measured.

4. How do electric and gravitational fields affect the motion of objects?

Electric and gravitational fields both exert forces on objects, causing them to accelerate and change their motion. Electric fields can cause charged particles to move in a certain direction, while gravitational fields can cause objects to orbit around a larger mass. The strength and direction of the fields determine how they will affect the motion of objects.

5. How do you solve problems involving electric and gravitational fields?

To solve problems involving electric and gravitational fields, you must first identify the given information and the unknown quantity you are trying to find. Then, use the appropriate equations and principles, such as Coulomb's law for electric fields and Newton's law of gravitation for gravitational fields, to solve for the unknown quantity. It is also important to pay attention to units and use proper mathematical techniques to ensure accurate solutions.

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