University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2024
Views:

248

Pages:

18
Author:

Dixie Summers

Electric Field Lines: “pictorial lines” Electric field lines help us visualize the electric field and predict how charged particles would respond to the field. Electric Field Lines Example: electric field lines for isolated +2e and -e charges. The number of lines leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge. Electric Flux We are now going to count the number of electric field lines passing through a surface, and use this count to determine the electric field. E The electric flux passing through a surface is the number of electric field lines that pass through it. Because electric field lines are drawn arbitrarily, we quantify electric flux like this: E=EA, A E …except that… If the surface is tilted with angle  , fewer lines cut the surface. Later we’ll learn about magnetic flux, which is why I will use the subscript E on electric flux. E  The green lines miss! We define A to be a vector having a magnitude equal to the area of the surface, in a direction normal to the surface. A  E  The “amount of surface” perpendicular to the electric field is A cos . Therefore, the amount of surface area effectively “cut through” by the electric field is A cos . AEffective = A cos  so E = EAEffective = EA cos . Remember the dot product from Physics 101? E  E  A If the electric field is not uniform, or the surface is not flat… divide the surface into infinitesimal surface elements and add the flux through each… A dA E  E  lim Ai  0  E  A i i  E   E  dA a surface integral, therefore a double integral Remember, the direction of dA is normal to the surface.  i If the surface is closed (completely encloses a volume), as the 6 faces of a cube … …we count lines going out as positive and lines going in as negative… E dA a surface integral, therefore a double integral  For a closed surface, dA is normal to the surface and always points away from the inside. Electric Flux Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction. E I see three parts to the cylinder: The left end cap. E dA Electric Flux Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction. E I see three parts to the cylinder: The tube. E Electric Flux Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction. E I see three parts to the cylinder: The right end cap. E dA Let’s separately calculate the contribution of each part to the flux, then add to get the total (net) flux. E   E  dA   E  dA left  left The left end cap.  E  dA tube tube   E  dA right right E dA The tube. The right end cap. E E dA E The left end cap. dA Every dA on the left end cap is antiparallel to E. The angle between the two vectors is 180  E  dA left left E is uniform, so   E dA left cos180  left  E dA left left  E dA left  E  dA left  EA left left left The tube. E E  Let’s look down the axis of the tube. E is pointing at you. dA Every dA is radial (perpendicular to the tube surface). E The angle between E and dA is 90. dA E The angle between E and dA is 90. dA  E  dA tube tube   E dA tube tube cos 90   0 dA tube 0 tube The tube contributes nothing to the flux! The right end cap. E dA Every dA on the right end cap is parallel to E. The angle between the two vectors is 0  E  dA right  right E is uniform, so  E dA right cos 0  right  right E dA right  E  E dA right right  right dA right  EA right The net (total) flux E   E  dA left left   E  dA tube tube   E  dA right right E  EAleft  0  EAright  0 The flux is zero! Every electric field line that goes in also goes out. E The net electric flux through a closed cylindrical surface is zero. + - E If there were a + charge inside the cylinder, there would be more lines going out than in. If there were a - charge inside the cylinder, there would be more lines going in than out… …which leads us to…