Potential Differences in a Uniform Electric Field

AI Thread Summary
The discussion revolves around a physics problem involving a block with an electric charge attached to a spring in a uniform electric field. The key equations used relate the potential energy of the spring and the electric field, leading to the expression for maximum spring expansion. Participants explore the conservation of energy principle, equating initial and final energies to derive the spring's maximum displacement. There is confusion regarding the signs of potential energy and work done by friction when introducing kinetic friction into the scenario. The conversation highlights the complexities of energy conservation in systems influenced by both electric fields and friction.
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Homework Statement



A 4.00 kg block carrying a charge Q = 50.o micro-C is connected to a spring for which k = 100 N/m. The block lies on a frictionless horizontal track, and the system is immersed in a uniform electric field of magnitude E = 5.00 X 10^5 V/m directed as shown in Figure P25.11. (not shown here, but the block is attached to a spring on it's left and the electric field is pointing to the right). If the block is released from rest when the spring is unstretched (at x = 0), (a) by what maximum amount does the spring expand?

Homework Equations



\Delta U = q_0 \Delta V = -q_0 E \cdot x
U_{spring} = \frac{1}{2}k x^2

The Attempt at a Solution



\frac{1}{2}k x^2 = Q E x
x = \frac {2 Q E}{k}

and that provides the correct solution but why does this work? I realize I'm setting the potentials as equals, but how can I do this?

We have (a) the change in potential energy from the field and (b) the change in potential energy from the spring. Since the spring is at rest at first, it's total Energy is 0, but when it reaches it's maximum, it has only potential energy. How can we equate this?

Or, can we say that it does have some energy at first, the potential energy from the electric field?

E_i = KE_{spring} + PE_{spring} + PE_{E field} = 0 + 0 + PE_{E field}

and then later we have:

E_f = KE_{spring} + PE_{spring} + PE_{E field} = 0 + PE_{spring} + 0

then by the law of conservation of energy, we may equate these? So,

E_i = Q E x = \frac{1}{2} k x^2 = E_f

?

I made a B last semester; thanks everyone for your help! :-)
 
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or, should I say that
W_{spring} + W_{E field} = 0
\frac{1}{2} k x^2 - Q E x = 0

when the total work is 0?
 
(d) Repeat part (a) if the coefficient of kinetic friction between block and surface is 0.200.

So I tried E_f - E_i = -f_k d
-E Q x - \frac{1}{2} k x^2 = - \mu m g x

but in this case the value for x would be non-real; twiddling with the signs failed to produce the correct answer. What am I not understanding correctly?
 
This seemed to do it:
E_f + E_i = -f_k x
where E_f = \frac{1}{2} k x^2 at the strongest point it's energy is the potential of the spring
and E_i = - E Q x at x = 0, the energy is the potential of the electric field on the charge

It seems odd that E_i is negative.
 
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