Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A small test charge is moved in the field of a fixed point charge . (a) The test charge is carried along a circular arc, with at the centre. How much work is done? (b) It is carried straight outwards from a distance to a distance , and then brought back to along a long, curved path. What is the net work? (c) A student draws an equipotential line that meets a field line at . What is wrong?
An equipotential surface has the same potential at every point, and no work is done along it. The field is perpendicular to it and points towards lower potential. The component of the field along any direction is , and for a small step .
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. The arc. Every point on a circle about is at the same distance from , so is the same at every point. The potential difference is zero, so the work is zero.
Step 2. Out and back. The field of a fixed charge is conservative. The work depends only on the end points. Going out and coming back to the same point means the same start and end point, so the net work is zero. The straight outward part is positive or negative, and the return path cancels it.
Step 3. The student. The field must be perpendicular to an equipotential line. If it met the line at , the field would have a part along the line. A charge moving along the line would then gain or lose energy, and the potential would not be constant along it.
Answer: zero work along the arc, zero net work for the closed path, and the line must cross the field line at .
Log in to join the discussion.
Loading…
With Standard, you also open 11 more problems, 2 Python labs and 1 extra course. Everything you have used here stays free, always.