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Problem-Solving Strategy Gauss’s Law IDENTIFY the relevant concepts: Gauss’s law is most useful in situations where the charge distribution has spherical or cylindrical symmetry or is distributed uniformly over a plane. In these situations we determine the direction of from the symmetry of the charge distribution. If we are given the charge distribution, we can use Gauss’s law to find the magnitude of Alternatively, if we are given the field, we can use Gauss’s law to determine the details of the charge distribution. In either case, begin your analysis by asking the question, “What is the symmetry?” SET UP the problem using the following steps: 1. Select the surface that you will use with Gauss’s law. We often call it a Gaussian surface. If you are trying to find the field at a particular point, then that point must lie on your Gaussian surface. 2. The Gaussian surface does not have to be a real physical surface, such as a surface of a solid body. Often the appropriate surface is an imaginary geometric surface; it may be in empty space, embedded in a solid body, or both. 3. Usually you can evaluate the integral in Gauss’s law (without using a computer) only if the Gaussian surface and the charge distribution have some symmetry property. If the charge distribution has cylindrical or spherical symmetry, choose the Gaussian surface to be a coaxial cylinder or a concentric sphere, respectively. EXECUTE the solution as follows: 1. Carry out the integral in Eq. (22.9). This may look like a daunting task, but the symmetry of the charge distribution and your careful choice of a Gaussian surface makes it straightforward. 2. Often you can think of the closed Gaussian surface as being made up of several separate surfaces, such as the sides and ends of a cylinder. The integral over the entire closed surface is always equal to the sum of the integrals over all the separate surfaces. Some of these integrals may be zero, as in points 4 and 5 below. 3. If is perpendicular (normal) at every point to a surface with area A, if it points outward from the interior of the surface, and if it also has the same magnitude at every point on the surface, then over that surface is equal to EA. If instead inward, then and and is perpendicular and 4. If is tangent to a surface at every point, then integral over that surface is zero. 5. If and the at every point on a surface, the integral is zero. 6. In the integral is always the perpendicular component of the total electric field at each point on the closed Gaussian surface. In general, this field may be caused partly by charges within the surface and partly by charges outside it. Even when there is no charge within the surface, the field at points on the Gaussian surface is not necessarily zero. In that case, however, the integral over the Gaussian surface—that is, the total electric flux through the Gaussian surface—is always zero. 7. Once you have evaluated the integral, use Eq. (22.9) to solve for your target variable. EVALUATE your answer: Often your result will be a function that describes how the magnitude of the electric field varies with position. Examine this function with a critical eye to see whether it makes sense.