Φ21 Fall 2006 HW6 Solutions

1 Problem 29.16

A proton with an initial speed of 800,000 m/s is brought to rest by an electric �eld. Part A. Did the protonmove into a region of higher potential or lower potential? Part B. What was the potential di�erence thatstopped the proton?

Solution: Since the kinetic energy decreased, the potential energy must have increased. The proton has apositive charge, so this is an increase in electric potential, i.e. moving to a region of higher potential.

Conservation of energy is used to say that the loss in kinetic energy equals the gain in potential energy. The�nal kinetic energy is zero while the initial kinetic energy is

K =12mv2 =

12

(1.67 × 10−27 kg

) (8.0 × 105m/s

)2= 5.344 × 10−16 J

The change in electric potential is then found by

5.344 × 10−16 J = −∆K = ∆U = q∆V

∆V =∆U

q=

5.344 × 10−16 J1.6 × 10−19 C

= 3340V

2 Problem 29.46

What is the ratio ∆Vp/∆Ve of the magnitudes of the potential di�erences that will accelerate a proton andan electron from rest to Part A. the same �nal speed and Part B. the same �nal kinetic energy?

Solution: To solve this problem, start with the requested expression and substitute for the variables untilonly the given information is in the expression. The required formulas are ∆V = q∆U , −∆K = ∆U , andK = 1

2mv2.

∆Vp

∆Ve=

qp∆Up

qe∆Ue=

e∆Up

−e∆Ue=

−e∆Kp

e∆Ke=

−e12mpv

2p

e12mev2

e

= −mp

me= −1833

(The book asks for the magnitude.) For part B, use the fourth fraction to say that |∆Vp/∆Ve| = 1.

3 Equipotential Surfaces in a Capacitor

Part A. Is the electric potential energy of a particle with charge q the same at all points on an equipotentialsurface?

Solution: Yes. That's what �equipotential� means.For a particle with charge q on an equipotential surfaceat potential V , the electric potential energy has a constant value qV .

Part B. What is the work required to move a charge around on an equipotential surface at potential V withconstant speed?

Solution: W = 0. Work changes the energy in a system. So Ei + W = Ef or ∆K + ∆U = W . Since thespeed of the charge is described as constant and it moves along the equipotential surface, both the kineticand potential energy are constant and there is no work.

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Part C. What is the work done by the electric �eld on a charge as it moves along an equipotential surfaceat potential V?

Solution: Work done by the electric �eld = 0. The work done by the electric �eld is equal to the change inthe electric potential energy, which is zero as the charge moves along the equipotential surface.

Part D. The work WE done by the electric �eld ~E in displacing a particle with charge q along the path ~dis given by

WE = q ~E · ~d = q| ~E||~d| cos θ

where theta is the angle between ~E and ~d. Since in general, ~E is not equal to zero, for points on anequipotential surface, what must θ be for WE to equal 0? Express your answer in radians.

Solution: θ = π/2rad. This shows equipotential surfaces are always perpendicular to the electric �eld.

Figure 1: Sketch of a capacitorand its ~E �eld direction.

Now assume that a parallel-plate capacitor is attached across the ter-minals of a battery as shown in the �gure. The electric �eld ~E in theregion between the plates points in the negative z direction, from higherto lower voltage.

Part E. Find the electric potential V (x, y, z) at a point ~X = (x, y, z)inside the capacitor if the origin of the coordinate system O = (0, 0, 0)is at potential 0. Express your answer in terms of some or all of thevariables E, x, y, and z.

Solution: V (x, y, z) = E · z is constant as x and y change an also has

the proper derivative to correspond to the given ~E �eld.

The equation of an equipotential surface at a potenial V0 is given by

z =V0

E

This is the equation of a plane that is parallel to the plates of the capacitor and perpendicular to the electric�eld. In particular, the lower plate, which is at zero potential, corresponds to the surface z = 0.

Part F. What is the distance ∆z between two surfaces separated by a potential di�erence ∆V ? Expressyour answer in terms of E and ∆V .

Solution: ∆z = ∆V/E derives directly from the solution to the previous part.

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4 Stopping the Proton

(Note: This problem has randomized numbers for each student.)

An in�nitely long line of charge has a linear charge density of 6.00×10−12 C/m. A proton is at distance 12.5cm from the line and is moving directly toward the line with speed 1200 m/s. How close does the protonget to the line of charge?

Solution: The electric �eld near an in�nite line charge is directed away from the positive charge with amagnitude calculated by Gauss's Law to be

E =λ

2πε0r

Since the force on the proton is non-uniform, the distance travelled is most easily calculated by conservationof energy.

∆K + ∆U = 0(0 − 1

2mP v2

i

)+ e∆V = 0

The electric potential ∆V is calculated by integrating the electric �eld.

∆V = −∫

~E · d~l

= −∫ rf

ri

λ

2πε0rdr

=−λ

2πε0ln

rf

ri

The distance of closest approach is

12mv2 = e

−λ

2πε0ln

rf

ri

−mv2πε0λ

= lnrf

ri

rf = ri exp−mv2πε0

eλ= 0.117m

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