Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 17

CHAPTER 17 1. We find the work done by an external agent from the work-energy principle: W = ®KE + ®PE = 0 + q(Vb – Va) = (– 8.6 × 10–6 C)(+ 75 V – 0)= – 6.5 × 10–4 J (done by the field).

2.

We find the work done by an external agent from the work-energy principle: W = ®KE + ®PE = 0 + q(Vb – Va) – 2.40 × 10–17 J (done by the field); = (1.60 × 10–19 C)[(– 50 V) – (+ 100 V)] = W = q(Vb – Va) – 150 eV. = (+ 1 e)[(– 50 V) – (+ 100 V)] =

3.

Because the total energy of the electron is conserved, we have ®KE + ®PE = 0, or ®KE = – q(VB – VA) = – (– 1.60 × 10–19 C)(21,000 V) = 3.4 × 10–15 J; 21 keV. ®KE = – (– 1 e)(21,000 V) =

4.

Because the total energy of the electron is conserved, we have ®KE + ®PE = 0; ®KE + q(VB – VA) = 0; 3.45 × 10–15 J + (– 1.60 × 10–19 C)(VB – VA); which gives VB – VA = Plate B is at the higher potential. 2.16 × 103 V.

5.

For the uniform electric field between two large, parallel plates, we have 4.2 × 104 V/m. E = ®V/d = (220 V)/(5.2 × 10–3 m) =

6.

For the uniform electric field between two large, parallel plates, we have E = ®V/d; 640 V/m = ®V/(11.0 × 10–3 m), which gives ®V = 7.04 V.

7.

Because the total energy of the helium nucleus is conserved, we have ®KE + ®PE = 0; ®KE + q(VB – VA) = 0; 65.0 keV + (+ 2e)(VB – VA); which gives VB – VA = – 32.5 kV.

8.

For the uniform electric field between two large, parallel plates, we have E = ®V/d; 3 × 106 V/m = (100 V)/d, which gives d = 3 × 10–5 m.

9.

We use the work-energy principle: W = ®KE + ®PE = ®KE + q(Vb – Va); 25.0 × 10–4 J = 4.82 × 10–4 J + (– 7.50 × 10–6 C)(Vb – Va), which gives Vb – Va = – 269V, or Va – Vb = 269 V.

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Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 17

10. The data given are the kinetic energies, so we find