Question

TRUE OR FALSE:

1. When the electric potential remains to be a constant along x axis (within a region), we know that the E field in the x direction is zero in that region.

2. The absolute value of the slope of the graph of V vs. x at any point (the derivative of V with respect to x) represents the magnitude of the E field along y and z direction at that point.

Think carefully, would the E fields in the y or z direction cause Potential change along the x direction?

3.  A continuous charge distribution can be considered as an infinite number of point charges dq, each dq creates a small amount of E field dE =ke dq /r2 at location P. To directly integrate the value of ke dq /r2 for all dq can find the total E field at that location.

4. A continuous charge distribution can be considered as an infinite number of point charges dq, each dq creates a small amount of E field dE at location P. Since E field is a vector, dE cannot be integrated directly for all dq to find the total E field at that location. dE must be decomposed into components in perpendicular directions (x or y directions for example) and only the components along the same direction can be integrated directly.

5. To calculate the potential created by a finite uniformly charged rod along the rod direction at a distance from one end of the rod, we integrate dV =ke dq /x, where x is the distance to dq. And dq is equal to the linear charge density times its length dx.

6. To calculate the E field created by a uniformly charged thin ring of radius a at one point along the ring’s axis with distance x from the ring’s center, we integrate dE =ke dq /r2, where r =sqrt(a2+x2) is the distance to dq. Because r is the same for all dq, the result will be equal to keQ /r2, where Q is the total charge, as if it is the E field created by a point total charge Q at a distance r.

7. To calculate the E field created by a uniformly charged thin ring of radius a at one point along the ring’s axis with distance x from the ring’s center, we consider dE created by each dq. The result’s magnitude will be less than that of ke Q /r2, where Q is the total charge. Because the charges spreads out on a ring, so the dE vectors due to those dq point to different directions and partially cancel out.

8. To calculate the potential created by a uniformly charged thin ring of radius a at one point along the ring’s axis with distance x from the ring’s center, we integrate dV =ke dq /r, where r =sqrt(a2+x2) is the distance to dq. The result will be equal to keQ /r, where Q is the total charge. Because r is the same for all dq and all of the potential contributions due to each dq directly add together, as if it is the potential created by total Q at a distance r.

9. A continuous charge distribution can be considered as an infinite number of point charge dq, each dq creates a small amount of potential dV =ke dq /r2 at location P. To directly integrate ke dq /r2 for all dq can find the total potential at that location. (Careful. Is potential proportional to charge over r square?)

10.  A continuous charge distribution can be considered as an infinite number of point charges dq, each dq creates a small amount of potential dV =ke dq /r at location P. To directly integrate ke dq /r for all dq can find the total potential at that location.

11. To calculate the potential created by a uniformly charged thin ring of radius a at one point along the ring’s axis with distance x from the ring’s center, we integrate dV =ke dq /r, where r =sqrt(a2+x2) is the distance to dq. The magnitude will be less than that of ke Q /r, where Q is the total charge. Because the charges spreads out on a ring, instead of concentrates as a point charge Q.

12. Integrating -dx/x2 from x1 to x2 gives 1/x23 – 1/x13.

13. Integrating dx/x from x1 to x2 gives 1/x22 – 1/x12.

14. Integrating -dx/x2 from x1 to x2 gives 1/x2 – 1/x1.

15.  Integrating dx/x from x1 to x2 gives ln(x2) - ln(x1).

Answer 1