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What happens to the size of the electric filed if the charge on the test charge is halved?
Wiki User
∙ 12y ago
Test charge is always a test charge. The electric field does not depend on the test charge. Usually we assume the test charge to be one coulomb positive charge. Though you make it half, it would never affect the field around the primary charge
Wiki User
∙ 12y ago
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Why did Nikola Tesla invent the fluorescent lighting?
Edison filed a suit that stated that Tesla could not use those for the electric fair in 1893 so he made his own.
The electric field inside a hollow uniformly charged sphere is zero Does this imply that the potential is zero inside the sphere Explain?
If the net charge enclosed by a surface is zero then the field at all points on the surface is not zero because gauss's law states that if the charge enclosed by a surface is zero then the flux through the surface is zero which depends upon the magnitude of field and the angle that it makes with the area vector at each point and so it is not necessary that the field will be zero at all points of the surface.
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It is only zero if there is no current in the conductor. When there is a current traveling through a long copper wire, there is an electric field helping the electrons over come the resistance. Because copper is a good conductor however, the electric field does not have to be very strong.That being said, if you have a piece of wire or any conductor and there is no battery or anything causing current to travel through it the electric field will always be zero inside.The easiest way to understand why is to imagine that there was an electric field in the conductor. If there was an electric field it would cause electrons to move, and moving of the electrons would result in the cancellation of the electric field.Here is a concrete example.Imagine a thick copper wire stretched from the left to the right. Image we were somehow able to pick electrons out of the left side and place them in the wire at the right end. For a very short moment after we placed the electrons in the right end there would be an electric field running through the entire length of the wire. The electric field would exert a force on every single electron in the copper wire and it would be toward the left. That is because they would be attracted to the exposed protons at the left end and be repelled by the lone electrons at the right end. Also, because of the electric field all the protons would feel a force towards the right because they would be attracted to the lone electrons at the right and repelled by the exposed protons at the left. The only charges that would move however would be the electrons that are free to move. As it happens, in copper, there is about one free electron per copper atom. Those free electrons would be shifted toward the left and the result would be that the exposed protons at the left would be "covered" up or neutralized and the lone electrons at the right would no longer be alone and therefor neutralized.To understand this process better imagine a long hose running from the left to the right and completely filled with marbles. Imagine pulling a marble out of the left end and holding it at or near the right end. The marble being held at the right end represents a lone electron and the empty space at the left end represents an exposed positive charge. Now if you were to to stick the marble into the right end, all of the marbles would shift to the left. As a result there would be no more empty space at the left end and no longer a lone marble at the right end.In the same way, because of the electric field in the copper wire, all the free electrons would shift to the left and the electric field would disappear because there would be no more exposed charges at either end. As it happens, this process would only take a fraction of a nanosecond.As a side note, what a battery does is force exposed positive charges to be at one end of a wire and lone electrons to be at the other. When the free electrons in the wire experience the electric field and shift, it is the "job" of the battery to replace the exposed positive charge and the lone electrons. In a normal circuit however there is a resistor at some point along the wire in order to reduce the rate of flow of electrons to something the battery can handle. The vast majority of the electric field/voltage ends up being across the resistor.So to recap, there is no electric field in a conductor because if there was, the free electrons would immediately shift to eliminate the electric field.At the risk of complicating matters I need to add that if the initial electric field in the conductor was not caused by charges in the conductor but by charges outside and distant from the conductor, the free electrons would still shift to neutralize the electric field but they would end up creating lone negative electrons and exposed positive protons at different places on the surface of the conductor to do it.The electric field created by the electrons shifting ends up exactly canceling the electric field created by the distant charges.A Wikipedia article that deals with this subject somewhat is in the related links section below.
What physical principle involving electric charges and a magnetic field is employed in the production of an electric current by an electric generator?
The physical principle that produces and electric current by an electric generator is theConservation of the Magnetic field: 0 =(d/dr + Del)(Bs + Bv)The field B is a quaternion field with the scalar part Bs and the vector part Bv.0= dBv/dr + Del Bs + DelxBv = dBv/cdt + Del Bs + DelxBvThis is the vector part (generator) of the Conservation,The magnetic field and electric field are related E=cB thus0= dBv/cdt + Del Bs + DelxBv = dBv/dt + Del Es + DelxEvFaraday did not include the scalar field and the term Del Es is missing from his Law.Del Es is the gradient of the scalar electric field and provides the back emf.The scalar part of the conservation of B is:0= dBs/dr - Del.BvHere too, the scalar filed Bs is left out and scientists declare that Del.Bv is always zero.This too is wrong dBs/dt = Del.Ev = rho/epsilon = zc rho = zJs
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