- #1

rbrayana123

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## Homework Statement

Consider two electrodes 2 mm apart in vacuum connected by a short wire. An alpha particle of charge 2e is emitted by the left plate and travels directly towards the right plate with constant speed 10

^{6}m/s and stops in this plate. Make a quantitative graph of the current in the connecting wire, plotting current against time. Do the same for an alpha particle that crosses the gap moving with the same speed but at an angle of 45°. Suppose we had a cylindrical arrangement of electrodes with alpha particles being emitted from a thin wire on the axis of a small cylindrical electrode. Would the current pulse have the same shape?

## Homework Equations

Uniqueness Theorem

For a cylinder, E = 2kλ/r

## The Attempt at a Solution

d = 2 * 10

^{-3}m

v = 10

^{6}m/s

Let Plate 1 be positioned at x = 0; point charge at x = a; and Plate 2 at x = d.

Also, Plate 3 will be positioned at x = a in different scenario;

E

_{1}is the electric field between Plates 1 & 3;

E

_{2}is the electric field between Plates 2 & 3.

Application of Uniqueness Theorem: If I find a solution for specific charges Q

_{1}on Plate 1 & Q

_{2}on Plate 2 given a point charge located at a, then it is THE solution. (Did I miss any points? I'm still a little iffy on this.)

Placing Plate 3 of total charge Q parallel to the other two plates and in the plane of x = a will distribute the same charges Q

_{1}& Q

_{2}as the a single point charge at x = a.

E

_{1}= σ

_{1}/ε

_{o}

E

_{2}= -σ

_{2}/ε

_{o}

Because they are connected by wires, the potentials are equal.

σ

_{1}/ε

_{o}(-a) = -σ

_{2}/ε

_{o}(d-a)

σ

_{2}= σ

_{1}a/(d-a)

Q

_{2}= Q

_{1}a/(d-a)

Q

_{1}+ Q

_{2}= -Q

Q

_{1}= -Q[itex]\frac{d-a}{d}[/itex]

Q

_{2}= -Q[itex]\frac{a}{d}[/itex]

Now, considering that there is a moving charge, a = vt, Q = 2e

Q

_{2}= -2e[itex]\frac{vt}{d}[/itex]

I = dQ/dt = -2ev/d

The graph of current versus time for a plate capacitor is constant. Now, if the charge were moving at a 45 degree angle, it's just a vector consideration.

I = dQ/dt = -2ev/d cos(45°) = -sqrt(2)ev/d

**Here's the hard part: The concentric cylinders**

Let the inner cylinder be of radius a with total charge Q

_{1}; and the outer cylinder be of radius b with total charge Q

_{2}.

Application of Uniqueness Theorem: If I find a solution for specific charges Q

_{1}on Cylinder 1 & Q

_{2}on Cylinder 2 given a point charge located at vt, then it is THE solution. Now, let me consider Cylinder 3 of radius vt with total charge Q to satisfy the above conditions.

E

_{1}= 2kλ

_{1}/r

E

_{2}= 2kλ

_{2}/r

Because they are connected by wires, the potentials are equal:

2kλ

_{1}ln(vt/a) = 2kλ

_{2}ln(b/vt)

Q

_{2}= Q

_{1}[itex]\frac{ln(vt/a)}{ln(b/vt)}[/itex]

Q

_{1}+ Q

_{2}= -2e

Q

_{1}(1 + [itex]\frac{ln(vt/a)}{ln(b/vt)}[/itex]) = -2e

Q

_{1}[itex]\frac{ln(vt/a) + ln(b/vt)}{ln(b/vt)}[/itex] = -2e

Q

_{1}[itex]\frac{ln(b/a)}{ln(b/vt)}[/itex] = -2e

Q

_{1}= -2e [itex]\frac{ln(b/vt)}{ln(b/a)}[/itex]

I = dQ

_{1}/dt = 2e[itex]\frac{1}{tln(b/a)}[/itex]

^This seems really off...