Non uniform charge density and electric potential

[usfelectrical]

As shown in the figure, a rod of length 9.8 m lies along the x-axis, with its left end at the origin. The rod has a non-uniform linear charge density λ = αx, where α = 0.009 C/m2 and x is the position. Point A lies on the x-axis a distance 3.59 m to the left of the rod, as shown in the figure.
Q is the total charge on the rod. The Coulomb constant is 8.988 × 109 N m2/C2.

Find the potentia at A, and answer in units of volts




the two relevant equations i can find are:
v=kq/r
dv=kdq/r and lambda=qx?


i am completely stuck on this problem and just need to find a way to start it. I though that you might be able to integrate the charge density formula over the total area of the rod and treat that value as Q and then plug that value into v=kq/r where r is the 3.59, but apparently that is worng and i am back to square one with no idea as to what to do

 

[ehild]

the two relevant equations i can find are:
v=kq/r
dv=kdq/r and lambda=qx?


i am completely stuck on this problem and just need to find a way to start it. I though that you might be able to integrate the charge density formula over the total area of the rod and treat that value as Q and then plug that value into v=kq/r where r is the 3.59, but apparently that is worng and i am back to square one with no idea as to what to do

The potential is additive, you need to integrate the contribution dV of a small segment of the rod dx with charge dq=λdx for the whole rod:
V(A)=∫(kdq/r)=∫(kλdx/r).

ehild

 

[usfelectrical]

wait so then what would be the values for the r? cause i have been using just 3.95 since that's the distance of your point away from the rod but it doesn't work

 

[Chestermiller]

wait so then what would be the values for the r? cause i have been using just 3.95 since that's the distance of your point away from the rod but it doesn't work

Maybe this will help: r = 3.59 + x

 

[usfelectrical]

ok, so then in the last integral you will have a ln, but would that be the same that you integrate over?

 

[Chestermiller]

As shown in the figure, a rod of length 9.8 m lies along the x-axis, with its left end at the origin. The rod has a non-uniform linear charge density λ = αx, where α = 0.009 C/m2 and x is the position. Point A lies on the x-axis a distance 3.59 m to the left of the rod, as shown in the figure.
Q is the total charge on the rod. The Coulomb constant is 8.988 × 109 N m2/C2.

Find the potentia at A, and answer in units of volts




the two relevant equations i can find are:
v=kq/r
dv=kdq/r and lambda=qx?

This equation is wrong. It should be dq = \lambdax dx.

Integrate this over the total length of the rod to get \lambda expressed in terms of q. Then integrate dv=kdq/r to get total v at A.

 

[usfelectrical]

i know that you have to integrate dq over the rod, what I'm saying is what do you integrate dv over?

 

[ehild]

dV is the contribution from dq to the potential at A (with respect to infinity). The integral of dV is simply V(A) the potential at A. Without charge, V is zero.

ehild

 

[usfelectrical]

ok, so the integral for dv would be from 0 to 3.59 because you're integrating to the potential at A?

 

[ehild]

dV is in volts, not in meters. dV is integrated over the potential belonging to zero charge to the actual potential V(A) of the total charge of the rod. The integral on the right-hand side goes along the length of the rod, from x= to x=9.8 m.
You need to use the 3.59 m in the term "r", distance from the charge dq to A.

ehild