Phys 202 1st Edition Lecture 4 Outline of Last Lecture I. Planar charge and mediumsOutline of Current Lecture II. Review From Physicsa. Trianglesb. ForcesIII. Solving with Coulombs lawCurrent LectureReview From Physics 201:Triangles: The sides of a right triangle can be measured using the equation a2+b2=c2 where c corresponds to the hypotenuse of the triangle. In the case of an equilateral triangle, we can simply divide thetriangle in half and measure the sides of the two right triangles this creates. The angle between the hypotenuse and either side of the triangle is represented by the symbol θ. Cos(θ) is equal tothe adjacent side divided by the hypotenuse. Sin(θ) is equal to the measure of the opposite sidedivided by the hypotenuse. Tan(θ) is equal to the opposite side divided by the measure of the adjacent side. Similarly, tan(θ) is also equal to sin(θ)/cos(θ).Forces:We can calculate multiple types of forces. Weight is equal to the mass of an object times the force of gravity(9.81N). We can also calculate normal force, force of tension, friction force, and electrostatic force. One of the easiest ways to measure force is to remember that opposite forces on a stationary object have to balance out. If two forces aren’t exactly opposite one another, for instance, if one force is off at an angle, we can use the x and y components of the angled force to solve for opposite forces. This is to say, we can make a hypothetical triangle, using the known force as one side of this triangle.Example:Imagine you are trying to calculate normal force on a block sitting on an inclined plane. Normal force is not really opposite weight.These 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 N A B mgBecause weight goes straight down, we can make a hypothetical triangle using the weight (mg) as the y component. Side A of this theoretical triangle will be equal to the normal force of the plane on the cube and side B will be equal to the force required to push the block up the ramp.Applying this to Physics 202:Electrostatics: Solving problems with Coulomb’s law:Imagine you have two pendulums share that are suspended from a shared pivot point. The weights at the end of the pendulums are conducting spheres with the same charge (q) and the same mass(m) of 10g. The length(L) of each pendulum is 140cm and the two spheres are separated by a distance(x) of 10cm. What is the charge of one sphere?θ Lmq X mqRemember that the equation for Coulombs law is;Fe=Kq1q2/r2However, in this case we are using x to measure distance. If we want to know the charge, we need to firstcalculate the force between the objects. Because we only need to know one of the charges, we can divide the pendulum system in half an solve using just one of the pendulums and spheres. θTx Ty TFe Fg=mgWe know from earlier physics that we can use x and y components to figure out the tension in the pendulum.1. Ty=Tcosθ=mg2. Tx=Tsinθ=Kq2/x2Tsinθ/Tcosθ=k(q2/x2)/mg => tanθ = k(q2/x2)/mgq=√(tanθmgx2/k)This derived equation is the one we will use to solve for the charge. So, we need to solve for tanθ. We can use the hypothetical triangle of the pendulum length and the distance between the spheres.θTy LT Mg x/2Tanθ of the right triangle derived from dividing the pendulum in half will be equal to tanθ of the hypothetical triangle derived from tension. Therefore, we can use the equation a2+b2=c2 to solve for tanθ.tanθ=opp/adj = (x/2)/√(L2-(x2/4)) => tan(θ)=(.1/2)/√(1.42-(.12/4)) => tanθ=.0118 Now that we have this value, we can just plug it into our derived equation for q to get the charge:q=√(tanθmgx2/k) => q=√(.0118*10*9.81*.12/8.99X109) =>