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Consider the path of a test charge that follows a line at right angles to the radial direction. We can calculate the work required to move the charge along this path in the usual manner. Since the force vector and path vector are at right angles, the dot product is zero along the entire path. Thus the total work from point A to point B is zero. The potential difference between point A and point B is the work done per charge along this path, which is zero. Thus the potential at point A is the same as the potential at point B. Since the electric potential along this path does not change, it is called an equipotential surface. For a point charge, equipotential surfaces are circular around the charge. This is because the electric field points in the radial direction at all points from the charge. Let's consider the implications of an equipotential. Along an equipotential surface there is no change in potential and thus no work is done on a test charge along this path. Thus if the test charge was somehow constrained to move only along this path and released at rest, it would remain at rest. From one equipotential surface to another, however, the potential is different, and work will be done on the charge by the electric field. Now if the charge can move freely in any direction, it will be accelerated from one potential surface to another by the electric field. Since the movement of electrical charges represents a current, this tells us that currents flow not along, but across equipotential surfaces For the Van de Graaff generator seen here, an electric field is created that points approximately radially away from the outer surface of the generator. Which of the images best depicts the equipotential surfaces for the Van de Graaff generator? Since the electric field is in the radial direction, the potential surfaces follow the surface of the sphere. Which way would you orient a long light bulb near the Van de Graaff generator in order for it to light up?
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