Question

The electric potential inside a 10.0 -m-long linear particle accelerator is given by \(V=\left(3000-5 x^{2} / \mathrm{m}^{2}\right) \mathrm{V},\) where \(x\) is the distance from the left plate along the accelerator tube, as shown in the figure. a) Determine an expression for the electric field along the accelerator tube. b) A proton is released (from rest) at \(x=4.00 \mathrm{~m} .\) Calculate the acceleration of the proton just after it is released. c) What is the impact speed of the proton when (and if) it collides with the plate?

Short answer

Answer: The impact speed of the proton when it collides with the plate is \(7.34 \times 10^{4}\,\text{m/s}\).

Step-by-step solution

Find the expression for Electric field

To find the electric field expression, differentiate the given electric potential expression, \(V(x)\), with respect to \(x\). The electric field \(E\) is related to the potential by: \(E=-\frac{dV}{dx}\) Differentiating \(V(x)=3000-5x^2\): \(\frac{dV}{dx}=-10x\) So, the electric field is: \(E(x)=10x\)

Calculate the acceleration of the proton

First, find the value of the electric field when \(x=4.00\,\text{m}\): \(E(4.00) = 10(4.00) = 40.0\,\text{N/C}\) Now, we have to find the force exerted on the proton. The equation for the force on a charged particle in an electric field is: \(F = qE\) where \(q\) is the charge of the proton, and \(E\) is the electric field. The charge of a proton is: \(q = 1.6 \times 10^{-19}\,\text{C}\) Substituting the values, we get: \(F = (1.6 \times 10^{-19})(40.0) = 6.4 \times 10^{-18}\,\text{N}\) Now we can find the acceleration \(a\) of the proton using Newton's second law (\(F = ma\)). We need to use the proton mass, which is \(m = 1.67 \times 10^{-27}\,\text{kg}\). So we have: \(a =\frac{F}{m} = \frac{6.4 \times 10^{-18}\,\text{N}}{1.67 \times 10^{-27}\,\text{kg}} = 3.83 \times 10^{9}\,\text{m/s}^2\)

Calculate the impact speed of the proton

To find the impact speed, we need to determine the work done by the electric force. We can use the work-energy theorem: \(W = \Delta K\) where \(W\) is the work done, and \(\Delta K\) is the change in kinetic energy. The work done by an electric force is given by: \(W = q\Delta V\) From the starting point \(x = 4.00\,\text{m}\), the proton will collide with the plate when \(x = 0\,\text{m}\). The change in potential is: \(\Delta V = V(0) - V(4.00) = (3000 - 5(0)^2) - (3000 - 5(4.00)^2) = 80\,\text{V}\) So, the work done by the electric force: \(W = (1.6 \times 10^{-19})(80) = 1.28 \times 10^{-17}\,\text{J}\) Now, we can find the change in kinetic energy: \(\Delta K= K_f - K_i\) Since the proton starts from rest, \(K_i = 0\). Therefore: \(\Delta K = K_f = W = 1.28 \times 10^{-17}\,\text{J}\) Finally, we can calculate the impact speed. The kinetic energy is also given by: \(K = \frac{1}{2}mv^2\) Solving for the speed \(v\), we get: \(v = \sqrt{\frac{2\Delta K}{m}} = \sqrt{\frac{2(1.28 \times 10^{-17}\,\text{J})}{1.67 \times 10^{-27}\,\text{kg}}} = 7.34 \times 10^4\,\text{m/s}\) So, the impact speed of the proton when it collides with the plate is \(7.34 \times 10^{4}\,\text{m/s}\).