Concept explainers
Suppose you have the uniformly charged cube in FIGURE Q24.1. Can you use symmetry alone to deduce the shape of the cube 's electric field? If so, sketch and describe the field shape. If not, why not?
If symmetry alone can be used to deduce the shape of the cube’s electric field.
Explanation of Solution
Given:
The following figure of uniformly charged cube is given.
Using Gauss’s law, we can deduce electric fields, particularly those with a high degree of symmetry, simply from the shape of the charge distribution.
The pattern of the electric field through the surface is like that the shape of the charged body, if the Gaussian surface matches the symmetry of the charge dispersion inside. For a consistently charged dispersion, the shape of the Gaussian surface coincides with the shape of the electric field. So, the shape of the electric field is cubical in nature for a consistently charged cube shape that is the electric field at each point on the face of a solid shape is comparative and its direction is perpendicular to the face cube. The sketch of the electric field through each face of the consistently charged cube is appeared as follows.
Conclusion:
We can use the symmetry alone to predict the shape of the electric field, the electric field in each face of the cube will have the same magnitude and perpendicular to that face.
Want to see more full solutions like this?
Chapter 24 Solutions
Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
Additional Science Textbook Solutions
Biology: Life on Earth (11th Edition)
Campbell Biology (11th Edition)
Chemistry: A Molecular Approach (4th Edition)
Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)
Chemistry
Microbiology: An Introduction
- In an isothermal process, you are told that heat is being added to the system. Which of the following is not true? (a) The pressure of the gas is decreasing. (b) Work is being done on the system. (c) The average kinetic energy of the particles is remaining constant. (d) The volume of the gas is increasing. (e) Work is being done by the system.arrow_forwardNo chatgpt pls will upvotearrow_forward8.114 CALC A Variable-Mass Raindrop. In a rocket-propul- sion problem the mass is variable. Another such problem is a rain- drop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is dp dv dm Fext = + dt dt dt = Suppose the mass of the raindrop depends on the distance x that it has fallen. Then m kx, where k is a constant, and dm/dt = kv. This gives, since Fext = mg, dv mg = m + v(kv) dt Or, dividing by k, dv xgx + v² dt This is a differential equation that has a solution of the form v = at, where a is the acceleration and is constant. Take the initial velocity of the raindrop to be zero. (a) Using the proposed solution for v, find the acceleration a. (b) Find the distance the raindrop has fallen in t = 3.00 s. (c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s. (For many more intriguing aspects of this problem, see K. S. Krane, American Journal of…arrow_forward
- 8.13 A 2.00-kg stone is sliding Figure E8.13 F (kN) to the right on a frictionless hori- zontal surface at 5.00 m/s when it is suddenly struck by an object that exerts a large horizontal force on it for a short period of 2.50 time. The graph in Fig. E8.13 shows the magnitude of this force as a function of time. (a) What impulse does this force exert on t (ms) 15.0 16.0 the stone? (b) Just after the force stops acting, find the magnitude and direction of the stone's velocity if the force acts (i) to the right or (ii) to the left.arrow_forwardPlease calculate the expectation value for E and the uncertainty in E for this wavefunction trapped in a simple harmonic oscillator potentialarrow_forwardIf an object that has a mass of 2m and moves with velocity v to the right collides with another mass of 1m that is moving with velocity v to the left, in which direction will the combined inelastic collision move?arrow_forward
- Please solve this questionarrow_forwardPlease solvearrow_forwardQuestions 68-70 Four hundred millilitres (mL) of a strong brine solution at room temperature was poured into a measuring cylinder (Figure 1). A piece of ice of mass 100 g was then gently placed in the brine solution and allowed to float freely (Figure 2). Changes in the surface level of the liquid in the cylinder were then observed until all the ice had melted. Assume that the densities of water, ice and the brine solution are 1000 kg m-3, 900 kg m3 and 1100 kg m3, respectively. 68 Figure 1 400 400 Figure 2 1m² = 1x10 mL After the ice was placed in the brine solution and before any of it had melted, the level of the brine solution was closest to 485 mL. B 490 mL. C 495 mL. Displaced volume by ice. D 500 mL. weight of ice 69 The level of the brine solution after all the ice had melted was A 490 mL B 495 mL D 1100kg/m² = 909 xious mis 70 Suppose water of the same volume and temperature had been used instead of the brine solution. In this case, by the time all the ice had melted, the…arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning