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My book "Concepts of Physics (Satish K. Gupta)" says:

The electric field of a charge is the space property by virtue of which the charge modifies the space around itself. As a result, if any charge is brought in the space around the charge, it experiences electrostatic force. It may be pointed out that the source charge does not experience any force due to electric field produced by it. The electric field due to a source charge has its own existence and is present even if there is no test charge to experience the force.

NEWTON'S LAW III: To every action there is always opposed an equal reaction: or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts. — Whatever draws or presses another is as much drawn or pressed by that other. If you press a stone with your finger, the finger is also pressed by the stone.

By reading the statements given in my book and the newton's third law, I got a question. If we press a stone with our finger, the finger is also pressed by the stone. If a source charge exerts force on the test charge, it should also feel the same force in opposite direction. So, won't it mean source charge is experiencing force due to its own electric field? If it is true, it is in contradiction with my text book statement that the source charge experiences no force due to electric field produced by it.

I don't know whether I have misunderstood anywhere, or (if I am not misunerstood) is it that source charge experiences force due to it's own electric field?

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  • $\begingroup$ If a charge A exerts a force on charge B, charge B exerts a force on charge A. The same way the earth attracts you and you attract the earth. $\endgroup$ – jinawee Jan 19 '14 at 13:29
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If a source charge exerts force on the test charge, it should also feel the same force in opposite direction. So, won't it mean source charge is experiencing force due to its own electric field?

There will be force on the source charge, but due to electric field of the test charge. If the test charge was not present, there would be no force on the source charge.

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This is because it is assumed that the test charge does not produce any electric field of its own and its magnitude is negligibly small, so it doesn't apply any force on the test charge.

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Your example of pressing a stone and your finger being pressed is a case where the forces concerned are contact forces where the participating surfaces are in direct contact with each other hence enabling them to exert a force on each other(i.e the action-reaction pair). Contact forces don't generate force fields. But electrostatic force is a non-contact force. In case of electrostatic force, the force on any charge is not due to its own field but due to the field of other charges placed in its vicinity.

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An important point is that test charge is just a concept, it has some strikingly impossible features : 1. A charge of 1 C and still a point 2. No field of its own

Now clearlt these features are abnormal and surreal. But we use test charges to define an electric field of any charged body because it simplifies the mathematical model very much. Since the test charge does not produce any field we can simply do calculations for field of desired charged body and since its charge is 1 C one variable is eliminated.

It would not be correct to use Newton's law in such an environment which is even theoritaclly impossible. However when you would do calculations for 2 charges A and B both would exert forces on each other due to the fields of one another and not their own.

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