- #1
wilywolie
- 5
- 1
- Homework Statement
- physics 2
- Relevant Equations
- Q=k.q1.q1/r2
can you help me with case (a)? ı couldn't do thatCutter Ketch said:I assume since you didn’t post anything you are happy with your answer for a
actually, ı am not sure for (b) that is why ı uploaded here, how can ı solve b or can you show me that. thank you.Cutter Ketch said:Actually, I’m starting to question whether b looks good. Gauss’ law is certainly true, but equating the integral to the field times the area assumes that the electric field is constant over the surface. In many problems you can make that assertion by symmetry, but is that true here?
Cutter Ketch said:Actually, I’m starting to question whether b looks good. Gauss’ law is certainly true, but equating the integral to the field times the area assumes that the electric field is constant over the surface. In many problems you can make that assertion by symmetry, but is that true here?
what about case (a)?etotheipi said:I am also unsure as to why an application of Gauss' law is necessary. It seems OP has used a spherical surface to determine the field strength at a point outside the sphere, which whilst being perfectly valid, reduces to a statement of Coulomb's law. Since any spherically symmetric distribution of charge is equivalent to a point charge at the centre (though granted, this can be shown with Gauss' law so might be why the problem-setter included it!).
wilywolie said:what about case (a)?
integral E.ds / is it true ?etotheipi said:What is the electric field inside a conducting material?
wilywolie said:integral E.ds / is it true ?
etotheipi said:I am also unsure as to why an application of Gauss' law is necessary. It seems OP has used a spherical surface to determine the field strength at a point outside the sphere, which whilst being perfectly valid, reduces to a statement of Coulomb's law. Since any spherically symmetric distribution of charge is equivalent to a point charge at the centre (though granted, this can be shown with Gauss' law so might be why the problem-setter included it!).
Cutter Ketch said:Maybe I jumped in too quickly here. I have no idea how to do this problem. One thing I am pretty sure of: it isn’t as simple as either his Gauss’ law solution or your equivalent Coulomb’s law assertion. The charges on the two spheres will attract. They will concentrate on the near faces. The electric field on the surfaces will not be uniform (limiting the utility of Gauss’ law) and they won’t act like a point source at the center (so no simple Coulomb’s law). In fact, you can say for absolute sure the field in the middle will be larger than these assumptions gives.
Gauss' Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is important because it allows us to calculate the electric field produced by a distribution of charges without having to directly measure the field at every point.
To use Gauss' Law correctly, you must ensure that the closed surface you are considering encloses the charge distribution you are interested in and that the electric field is constant and perpendicular to the surface at every point. You must also make sure to use the correct form of the law for the given situation.
Gauss' Law has many practical applications, such as calculating the electric field inside a charged capacitor, determining the electric field of a point charge, and analyzing the electric field of a charged conducting sphere.
To solve problems using Gauss' Law, you must first identify the charge distribution and the closed surface you will use to apply the law. Then, you can use the appropriate form of the law to calculate the electric field at a specific point or over a given region.
Yes, Gauss' Law can be used for any charge distribution, as long as the electric field is constant and perpendicular to the surface at every point. However, it is often easier to apply the law for symmetric charge distributions, such as a point charge or a charged sphere.