How Can an Electron Microscope Manipulate an Electron Beam in 2D Motion?

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An electron microscope utilizes an electron beam to visualize extremely small objects, such as viruses, which are not detectable by light microscopes. The manipulation of the electron beam is achieved through electric and magnetic fields that act as lenses. In a specific example, an electron traveling along the x-axis experiences acceleration in both x and y directions, requiring calculations to determine its position at a given point. The initial confusion regarding the notation for initial velocity was clarified, leading to the successful application of kinematic equations to find the electron's velocity and time. Ultimately, the problem was resolved by correctly interpreting the equations and applying them to find the electron's position.
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Homework Statement


It is not possible to see very small objects, such as viruses, using an ordinary light microscope. An electron microscope can view such objects using an electron beam instead of a light beam. Electron microscopy has proved invaluable for investigations of viruses, cell membranes and subcellular
structures, bacterial surfaces, visual receptors, chloroplasts, and the contractile properties of muscles. The “lenses” of an electron microscope consist of electric and magnetic fields that control the electron beam. As an example of the manipulation of an electron beam, consider an electron traveling away from the origin along the x-axis in the xy plane with initial velocity \mathbf{v_i} = v_i \hat{i} . As it passes through the region x = 0 to x = d, the electron experiences acceleration \mathbf{a} = a_x \hat{i} + a_y \hat{j},where a_x and a_y are constants. For the case v_i = 1.80 \times 10000000 m/s, a_x = 8.00 \times 100000000000000 m/s^2 and a_y = 1.60 \times 1015 m/s^2, determine
at x = d = 0.0100 m the position of the electron

Homework Equations


x_f = x_i + v_xi t + .5 a_x t^2

The Attempt at a Solution


I can't seem to find the value of t. I've tried reorganizing equations I know that have t in them, but I can't get a value that, when plugged into the position as a function of time equation, makes sense. What am I missing?
 
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x_f = x_i + v_xi t + .5 a_x t^2

What is the "i" after v_x
? Should it be: v_{xi} as meant to be initial velocity in x-direction?

If you write more on how you tried to solve for t, we can help you more.

But are you sure how to solve a quadratic equation? Is that your problem?

at^{2} + bt + c = 0 ; t = - \frac{b}{a2} \pm \sqrt{(\frac{b}{a2})^{2} - c/2}
 
I figured it out

Never mind! I figured out the answer. To answer the above question, i just made a mistake in my latex code; the i should be subscripted as it represents initial. If your interested in how I found the answer, I found the velocity of the electron at d using the equation v^2_{xf} = v^2_{xi} + 2 a_x (x_f - x_i) and then plugging that value into the equation v_{xf} = v_{xi} + a_x t and solving for t. From there the problem is very easy.
 
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