1

The first thing that we should notice is that, due to the fact that the electric field is only in the x-direction, all motion in the y-direction is irrelevant to calculating the change in potential. Therefore, all we need to concern ourselves with is the change in position in the x-direction. The potential difference through which the charge has moved is given by

V = -E(x_f - x_i)

since E is constant. This gives a change in

V = -(250 V/m)(0.2 m) = -50 V.

Using this change in V, we get a change in potential energy of

qV = (12 x 10^-6 C)(-50 V) = -6.0 x 10^-4 J.

2

(a) If the proton has accelerated through 120 V, then it has "fallen" through a potential energy of

(120 V)(1.6 x 10^-19 C) = 1.9 x 10^-17 J.

If we assume that all of the potential energy change is converted to kinetic energy, then we know that

KE = .5 mv² = 1.9 x 10^-17 J.

Solving for v, we get

v = ((3.8 x 10^-17 J)/(1.7 x 10^-27 kg)) ^1/2 = 1.5 x 10⁵ m/s.

(b) Since the electron has the same charge on it as a proton, the potential energy change is the same. However, since the mass is different, we get

v = ((3.8 x 10^-17 J)/(9.1 x 10^-31 kg)) ^1/2 = 6.5 x 10⁶ m/s.

3

The energy of the electron is given by qV. Since the answer is requested in electron-Volts and since q is equal to the charge on an electron, we get that the energy is

(1 e)(20,000 V) = 20,000 eV.

This energy correspond to

20,000 eV = (1.6 x 10^-19 C)(20000 V) = 3.2 x 10 ^-15 J. If this is all converted to kinetic energy, then the velocity of the electron is given by

v = ((6.4 x 10^-15 J)/(9.1 x 10^-31 kg)) ^1/2 = 8.4 x 10⁷ m/s.

4

In order to calculate how much energy would be expended in moving the charge, we must first calculate the potential at the location where the charge is. This potential is simply the sum of the potentials due to the other two point charges at the location of the 8 microcoulomb charge. The potential at that point is

V = (9 x 10⁹ Nm²/C²)(2 x 10^-6 C)/(.03 m) + (9 x 10⁹ Nm²/C ² )(4 x 10^-6 C)/(.03² + .06²) ^1/2 m = 6 x 10⁵ V + 5.4 x 10⁵ V = 11.4 x 10 ⁵ V.

The potential of the two charges at infinity is 0. Therefore, the energy necessary to move the charge from its position on the rectangle to infinity is

0 J - (8 x 10^-6 C)(1.14 x 10⁶ V) = -9.1 J.

5

For a parallel plate capacitor, we know that the capacitance is given by

C = A/(4 pi k d).

==> A = 4 pi k d C = (4)(3.14)(9 x 10⁹ Nm²/C²)(1 x 10^-4 m)(2.0 x 10^-12 F) = 2.3 x 10^-5 m².

6

(a) If the capacitors are connected in parallel, then we know that the total capacitance is just the sum of the individual capacitances, i.e.

C_Total = C₁ + C₂ + C₃ = (5.0 + 4.0 + 9.0) x 10^-6 F = 1.8 x 10 ^-5 F. (b) If the capacitors are connected in series, then the total capacitance is just the reciprocal of the sum of the reciprocals, i.e.

1/C_T = 1/C₁ + 1/C₂ + 1/C₃ = (1/5.0 + 1/4.0 + 1/9.0) x 10⁶ 1/F = 5.6 x 10⁵ 1/F.

By inverting this, we get

C_Total = 1.8 x 10^-6 F.

7

The first thing that we need to do is to calculate the capacitance of the capacitor. After we find this, then we know that the energy stored on the capacitor is .5 CV². Using the data given, we have

C = A/(4 pi k d) = (2.0 x 10^-4 m²)/((12.6)(9 x 10⁹ Nm²/C² )(.005 m)) = 3.5 x 10^-13 C.

For a 12 V battery, we get that the stored energy is

(.5)(3.5 x 10^-13 F)(12 V) ² = 2.5 x 10^-11 J.

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If you find an error in these, or just have a comment, send an e-mail message to jpratte@kennesaw.edu