y (cm) -5 10 24 0 0.25 0.5 0.5 -0.25 -0.5 -10 -10 -5 0 5 10 x (cm) =(-0.8,4.6) cm V₁ = 0.34 V =(-0.8,4.6) cm V₂ = 0.34 V In this contour plot you can see lines of equal potential. The lines are labeled with their potential values in units of Volts. These potential lines are generated by a collection of point charges. Part 1: What is the minimum number of point charges that are needed to generate a potential surface like the one shown here? 3 You are correct. Your receipt no is 154-6458 ② Previous Tries Part 2: What is the value of the x-component of the electric field at the blue point with coordinates (-2.8,-1.7) cm? (Hint: you need to extract this value graphically. In order to do this, it helps to draw a straight horizontal line from one equipotential line to another, crossing the point. For your convenience, you can do this in the contour plot above by clicking on the starting point of your line, dragging the mouse, and then releasing at the endpoint. The intial and final coordinates of your line will be displayed below the contour plot. Once drawn, you can move the starting point of the line by using the arrow keys on your keyboard; holding the shift key down at the same time will move the endpoint of the line.) -50 V/m HINT: The electric field is the negative of the gradient of the electric potential. The endpoints of the line that you draw appear below the contour plot, as well as the values of the potential at these points. This way you can draw a line from any point to any other point, and you will know the value of the potential at both endpoints of the line. And the closer you get with the endpoints of your line to the point for which you want to determine the E-field, the better this approximation becomes. Submit Answer Incorrect. Tries 1/99 Previous Tries Part 3: What is the value of the y-component of the electric field at the same point? Submit Answer Tries 0/99 The point of the exercise in parts 2 and 3 is to let you see how the component of the electric field are related to the partial derivatives of the potential. Numerically, you compute a partial derivative with respect to x by leaving the other coodinate(s) constant and taking AV/Ax. For reasonably small values of Ax this should converge to av/ax. But, careful: if you make your interval too small, you may encounter large rounding errors. One more point: if you keep your blue point in the middle of your interval, you get faster convergence than when you select it as one of your endpoints. Post Discussion Send Feedback