A proton is launched from a very long negatively charged plate at an initial velocity \(v_o\) and angle \(\theta\) toward an identical, positively charged plate, as shown below. The two plates are held at a potential difference \(V=1000\) volts, and the distance between the two plates is \(d\). a) Draw in the electric field vectors. Write an algebraic expression for the electric field strength in terms of given quantities. b) If \(v_o=4 \times 10^5 \mathrm{~m} / \mathrm{s}\), and \(\theta=60^{\circ}\), does \(--_{-}--_{-}--_{-}--^{-}\) the proton hit the top plate? If so, with what velocity? c) If \(d=25 \mathrm{~cm}\), how long is the proton in flight? d) If a uniform magnetic field pointing along the proton's initial velocity vector is introduced between the plates, does this alter the conclusion to (b)?

The electric field vectors between the two plates are perpendicular to the plates and directed from the positively charged plate to the negatively charged plate. The electric field strength between the plates can be found using the formula: $$ E = \frac{V}{d} $$ Where \(E\) is the electric field strength, \(V\) is the potential difference between the plates (in this case, 1000 V), and \(d\) is the distance between the plates.

Let's find out whether or not the proton hits the top plate and, if it does, determine its final velocity. First, we need to separate the initial velocity into its horizontal and vertical components: $$ v_{0x} = v_{0} \cos{\theta} $$ and $$ v_{0y} = v_{0} \sin{\theta} $$ Given \(v_{0} = 4 \times 10^5\) m/s and \(\theta = 60^{\circ}\), $$ v_{0x} = 4 \times 10^5 \cos{60^{\circ}} \approx 2 \times 10^5 \mathrm{~m/s} $$ $$ v_{0y} = 4 \times 10^5 \sin{60^{\circ}} \approx 3.464 \times 10^5 \mathrm{~m/s} $$ Now, let's find the acceleration acting on the proton due to the electric field. Since the proton is a positively charged particle, it will experience an upward force due to the electric field: $$ F_{e} = qE $$ Where \(F_{e}\) is the electric force on the proton, \(q\) is the charge of the proton, and \(E\) is the electric field strength. Therefore, the acceleration experienced by the proton in the vertical (y) direction will be: $$ a_{y} = \frac{F_{e}}{m} = \frac{qE}{m} $$ Where \(m\) is the mass of the proton. Let's find the time it takes for the proton to hit the top plate. To do this, we will consider the vertical motion of the proton and use the equation: $$ y = v_{0y}t + \frac{1}{2}a_{y}t^2 $$ Since the proton moves from the bottom plate to the top plate, the vertical displacement \(y = d\). To find the time \(t\), we'll need to solve the above equation for \(t\). If the proton does hit the top plate, the time it takes will be a real positive value. The final velocity in the y-direction will be: $$ v_{fy} = v_{0y} + a_{y} t $$

Given \(d = 25 \mathrm{~cm} = 0.25 \mathrm{~m}\), we can calculate the time of flight using the equation we established in Part b. First, substitute the values and solve for \(t\), then evaluate the final vertical velocity using the time of flight we found.

If a uniform magnetic field is introduced along the direction of the proton's initial velocity vector (parallel to the x-axis), it will not directly impact the vertical displacement of the proton, as it's orthogonal to the electric field. However, it does change the motion of the proton by causing it to experience a magnetic force in addition to the electric force. Nonetheless, the magnetic force will not change the answer to part b, as it will not impact the vertical motion of the proton. The student should work through the calculations for each part using the provided equations and the given values for \(V\), \(d\), \(v_{0}\), and \(\theta\).