answer in m/s, A pair of charged conducting plates produces a uniform...

Question

A pair of charged conducting plates produces a uniform field of 12,000 N/C, directed to the right, between the plates. The separation of the plates is 40 mm. An electron is projected from plate A on the left, directly toward plate B on the right, with an initial speed of v0 = 2E7 m/s. What is the speed of the electron as it strikes plate B?  

Answer 1

Hi Anna A.!

There are several solutions for this answering this question. One way we can think about this is by using the work-energy theorem:

W = Fd = ΔKE

From here, we consider how the electric field strength, electrostatic force and charge are all related:

=> E = F_q*q^-1 => F_q = qE

So, we can apply the equation above and substitute it into our work-energy theorem:

W = Fd = ΔKE

=> E = F_q*q^-1 => F_q = qE

W = qEd = ΔKE

Now, we just rearrange, plug in and solve:

W = qEd = ΔKE ----> qEd = (1/2)m(v_f² - v_o²)

2qEd/m = v_f² - v_o² => v_f = √((2qEd/m) + v_o²)

<=> √[2*(-1.6x10^-19 C)(12x10³ N*C^-1)(4x10^-2m)/(9.1x10^-31kg) + (2x10⁷m*s^-1)²]

v_f = 1.5x10⁷ m*s^-1

As we discussed above, there are other solutions to this problem. For instance, we could have obtained the same final equation using potential energy:

ΔU_q = q*ΔV

Because the force here is one that is conservative, we can equate the potential energy to the kinetic energy:

ΔU_q = q*ΔV = ΔKE

Now, we can substitute for the difference in potential:

ΔV = Ed

If we plug this into our equation, we obtain the same equation as that above:

ΔU_q = q*ΔV = ΔKE

=> ΔV = Ed

ΔU_q = q*Ed = ΔKE -------> v_f = √((2qEd/m) + v_o²)

Still, there are more solutions possible. But I think I'll leave it here!

Hope this helps!

Cheers