Phys 202 1st Edition Lecture 2Outline of Last Lecture I. Coulombs Law and Electric FieldOutline of Current Lecture I. Coulomb’s LawII. Electric Field and CoulombsIII. FluxIV. Gauss’s LawCurrent LectureCoulombs Law Recap: If there were two point charges (q1 and q2) separated by a distance r, the point q2 experiences a force (F) due to q1 that can be expressed as;F=(1/4πEo)(q1q2/r2)Where q1 represents the quantity of charge on object 1 in Coulombs(C), q2 represents the quantity of charge on object 2, and r represents the distance between the two objects (in meters). If q1q2>0 then the charges are the same, meaning they repel. If q1q2<0, then one, but not both, are negative, meaning the charges attract. Electric force crucially depends on distance. The greater the distance r, the greater the force.Example:Two balloons are charged with an identical quantity and type of charge: -6.25 nC. They are held apart at a separation distance of 61.7 cm. Determine the magnitude of the electrical force of repulsion between them.First thing we have to do is convert our units into ones we can use. So, we convert -6.25nC to -6.25X10-9Cand we convert 61.7cm to .617m.Next we use the equation;F=k(q1q2)/r2Remember that k is equal to 8.89X109 Nm2/C2. So, F=(8.89X109 Nm2/C2)(-6.25X10-9C)(-6.25X10-9C)/(.617m)2These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.F=9.23X10-7NBecause the charges on the balloons were both negative, the magnitude of the force is positive, and therefore the force is repulsive. Electric Field with Coulombs law:The strength of an electric field is dependent on the amount of charge that makes the field, and the distance from the charge.Electric field can be calculated using an equation really similar to Coulombs law. The main difference is that instead of q1 and q2, we use Q to represent a source charge and q to represent the test charge. The source charge (Q) is the source of the electric charge. The magnitude of the electric field of the source charge can be measured using any other charge placed within the field. The test a single point chosen to measure the charge. The standard equation for the magnitude of electric field(E) is;E=F/qWhere F is the force per charge on the test charge q. Assuming you don’t have a value for F, you can substitute Coulombs law in for F (replacing q1 and q2 with Q and q). In this case, q in the equation for force and q in the denominator cancel out. So you are left with the equation;E=kQ/r2The stronger the charge, the more dense the electric field lines.FluxElectric flux(φ) is the rate at which an electric field flows through a given area. The flux is proportional to the number of electric field lines going through a hypothetical surface. If the electric field is uniform, then the electric flux passing through a surface of vector area A is Δφ=|E||ΔA|Cosθwhere E is the electric field, A is the area of the surface, and θ is the angle between the electric field lines and the normal (perpendicular) to A. You can kind of think of flux as the number of electric field lines that cross through a surface. The unit of flux is Weber.To find flux through all areas of a closed surface, you’d have to sum up all of the values of φ for an exceptionally large number of small sections of the surface. This sum of a constant flux is the integral of the surface.Gauss’s Law:Gauss’s law simplifies the task of calculating the flux through a closed surface. This law states that that the net electric flux through any closed surface is equal to the charge inside that surface divided by the constant epsilon (ε). The equation for this is;φelectric=Q/εPut another way, the total electric flux out of a closed surface is equal to the charge enclosed by the surface, divided by the permittivity. Since we know the equation for flux (φ), we can expand this equation