- #1
StephenDoty
- 265
- 0
If a rod with charge density lambda is bent into a 3/4 circle what is the E field at the center?
Well if the rod is 3/4 of a circle then neither the x or y components can cancel out, thus
E= k*lambda/ R (i) + k*lambda/R (j)
right?
Well if the rod is 3/4 of a circle then neither the x or y components can cancel out, thus
E= k*lambda/ R (i) + k*lambda/R (j)
right?