![]() Solution: If we want to use Gauss's law to find the flux for such a configuration, then we should construct a closed surface around the given charge. electric flux: volt metre: Vm kgm 3 s 3 A 1: E: electric field strength. If a net charge is contained inside a closed surface, the total flux through the surface is proportional to the enclosed charge, positive if it is positive. Notice that the unit of electric flux is a volt-time a meter. Solution: The electric flux which is passing through the surface is given by the equation as: E E.A EA cos. ![]() E E is the magnitude of the electric field at a point in space, k k is the universal Coulomb constant k 8.99 ×109Nm2 C2 k 8.99 × 10 9 N m 2 C 2, q q is the charge of the particle. Find the electric flux that passes through the surface. Find the magnitude of the electric flux that passes through the square. Electric charges attract or repel one another with a force inversely proportional to the square of the distance between them. In equation form, Coulomb’s Law for the magnitude of the electric field due to a point charge reads. Correct Option: (d) Explanation: The formula for electric flux becomes the formula E.s.cos because both the area and electric field are vector quantities, and we can also express electric flux as E. Problem (medium): An electron is placed at a distance of $\ell/2$ just above the center of a square of edge $\ell$. 4) If the angle between the field lines and area vector is, the electric flux will be maximum. ![]() In simple words, Gauss's law states that the net number of electric field lines leaving out of any closed surface is proportional to the net electric charge $q_\] Gauss's Law Problems and Solutions Gauss's Law Definition: What is the net electric flux on the spherical surface between the charges Briefly explain your answer. When we need to use Gausss law for any problem, and then evaluate the electric flux through a chosen Gaussian surface, we only (need to) consider the (net).
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