- #1
Irishdoug
- 102
- 16
- Homework Statement
- Please see picture as diagram required for answer. I am looking to see if my answers are correct as I am unsure if they are. Thankyou for your help.
- Relevant Equations
- The E-field in the x-direction is zero for the charge at y due to symmetry , and in the y-direction is negative.
W = F*scos##\theta##
a) From X -Y. The work done on the positive charge is negative as the displacement is in the negative y-direction i.e. It is a positive charge moving in parallel to a negative E-field: W= F*(-s) = (+)(-) = -
b) Y-Z. The work done is 0. The E-field in the x-direction is 0 as they cancel due to symmetry. ##E_{y}## is also 0 as the charge is moving perpendicular to the Electric field in the y-direction. That is, Fscos##\theta## = Fscos(##\frac{pi}{2})## = 0. As there is no E-field there is no force to work against and the work done is 0.
c) E-potential for Y is greater than Z.
Y is equi-distant from both charges. As such, ##Vy##=##\frac{2kq}{r}##. The potential for Z in the x-direction --> ##Vz_{x}## = ##\frac{kq}{r}## cos##\theta##.
cos##\theta## = x/r . So ##Vz_{x}## = ##\frac{kqx}{r^{2}}##
Likewise ##Vz_{y}## = ##\frac{kqy}{r^{2}}##
As such: ##Vz_{x}## + ##Vz_{y}## < ##Vy## due to ##r^{2}## terms in the ##Vz## equation
d) It is positive. The new charge has an E-field twice as large as the current E-field. As such, we are now moving a positive charge in a positive E-field (that is moving a positive charge in the negative y-direction in an E-field pointing in the positive y-direction). Therefore we lose energy and do positive work.