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(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable. (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart. |
Answer» Solution :(a) For any electrostatic field configuration, test charge `q_0` (extremely small, point like positive charge) when placed at null point (where resultant electric field is zero) remains certainly under unstable equilibrium condition. Let us prove this by method of contradiction.  Suppose point 0 is null point and suppose test charge `q_0` is under stable equilibrium condition at that point. If really it is so then when `q_0`is displaced in any direction like `vec(OA), vec(OB)` or `vec(OC)`...... and then released, it should return back to same point O. For this to happen, restoring force MUST act towards null point O from all the directions for which all the electric field lines at point 0 should be radially inward. This can happen if Gaussian SURFACE imagined around point O encloses some negative charge ! But actually it is not so. Hence, our assumption is wrong. This means that test charge placed at null point remains under unstable equilibrium condition, (b) Consider a system of two identical point charges q and q lying respectively at POINTS A and B.  For above configuration (arrangement) null point is C which is midpoint of `bar(AB)`. Now, when test charge `q_0` is displaced from C to C. resultant Coulombian force on `q_0` is `2Fcostheta`, which is pointing away from null point C, which means that test charge `q_(0)` will never come back to its equilibrium POSITION at C. Thus, test charge qQ placed at C is under unstable equilibrium condition (because it had been stable equilibrium condition then test charge would have returned to position C from all the directions). |
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