How Much Work Is Needed to Move a Charge to the Origin from 1.25m?

[pyroknife]

Homework Statement

A charge of 9 nC is uniformly distributed around a ring of radius 10 cm that has its center at the origin and its axis along the x axis. A point charge of 1 nC is located at x = 1.25 m. Find the work required to move the point charge to the origin. Give your answer in both joules and electron volts.

Homework Equations

since it's a ring we can get an approximate answer if we treat the ring as a point charge at the origin or we can use v=kQ/(z^2+a^2)^.5 where z=1.25 a=.1
W=q*v

The Attempt at a Solution

w=(10^-9)(9x10^9)(9x10^-9)/(1.25^2+.1^2)^.5

which give me 6.5e-8 but the answer wasn't that.

 

[ehild]

You calculated the potential energy of the charge with respect to infinity. That is equal to the work required to bring the charge from infinity (where the potential energy is zero) to x=1.25 m, but the question is the work required to move it from x=1.25 m to x=0.

ehild

 

[pyroknife]

You calculated the potential energy of the charge with respect to infinity. That is equal to the work required to bring the charge from infinity (where the potential energy is zero) to x=1.25 m, but the question is the work required to move it from x=1.25 m to x=0.

ehild

do I have to do dU=Vdq where V is kQ/(z^2+a^2)^.5 and integrate that which gives me kQ^2/(2(z^2+a^2)^.5). I don't think that's right tho. If it is what would be my limits of intregration?

 

[ehild]

The work W required to move a charge from point A to point B is equal to the potential difference multiplied by the charge. W=q(U(B)-U(A)).
What is the potential at the origin, x=0?

ehild

 

[pyroknife]

wait isn't W=q(V(B)-V(A)? The above equation isn't in joules or is U electric potential and not potential energy?

wouldn't V(B) be 0 tho the electric potential at x=0 when the charge is moved there?

 

[pyroknife]

okay i got the right answer by doing w=F*dx. I integrated from x=0 to x=1.75. But I don't understand how that got me the right answer. Shouldn't you integrate from 1.75 to 0 since your starting point is at 1.75 not 0?

 

[ehild]

Some people use U for the potential, others use V. Let it be V.
The force required to move the charge is opposite to the electric force. When you integrate the electric force from zero to 1.25 m, it is the same as integrating the opposite force from 1,25 to 0.

The potential at a point A is defined as the negative work done by the electric field when a unit positive charge moves from the reference point P₀,where the potential is zero, to A.

$V(A)=-\int_{P_0}^A{\vec{E} d\vec{r}}$

In case of a point charge we choose the potential zero at infinity. The same holds for a ring of charge: The potential along the axis is V(x)=kQ/(x^2+a^2)^0.5, and it is zero at the limit when x tends to infinity.

If the zero of the potential is at infinity

$V(A)=\int_A^\infty{\vec{E} d\vec{r}}$.

As the integration is additive,

$V(A)=\int_A^B{\vec{E} d\vec{r}}+\int_B^\infty{\vec{E} d\vec{r}}$,

$\int_A^B{\vec{E} d\vec{r}}=V(A)-V(B)=-\Delta V$

The work done by the electric field when a charge q moves
from point A to B is equal to the integral of the electric force, qE :

$W(AB)=q\int_A^B{\vec{E} d\vec{r}}=q[V(A)-V(B)]=-q\Delta V$,

the same as the negative potential difference multiplied by q.

The work required to move the charge from A to B, done some external force against the electric field, is just the opposite, q[V(B)-V(A)].

Your potential is kQ/(x^2+a^2)^0.5, and the work required to move the charge q from x=1.25 m to x=0 is

$W=kqQ(\frac{1}{a}-\frac{1}{\sqrt{1.25^2+a^2}})$.