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Chapter 16 Problem 1 The magnitude of the work is given by the product |W | = qE∆x cos θ = 6.4 · 10−19 J. The angle is zero because the motion of the proton is parallel to the field. The proton moving along a field loses potential energy, thus the change in potential energy is also 6.4 · 10−19 J. The potential difference is V = E∆x = 4V Problem 3 The work required is going to be W = qE∆x = q∆V . Here, a sodium ion has a total charge of +1e = 1.6 · 10−19 C, and ∆V = 90mV . Therefore the work required is 90mV · 1.6 · 10−19 C = 1.44 · 10−20 J. Problem 11 The electric potential from a single charge is given by V = kre q where r is the distance from the charge. Plugging in r = 0.01m gives V = 1.44 · 10−7 V . The potential difference between two points, ∆V = V2 − V1 , is found by evaluating hte potential at each of the two points, and subtracting them. The first point has already been evaluated, the potential at the second point is 7.2 · 10−8 V , thus the difference is −7.2 · 10−8 V , negative because V1 is bigger than V2 . Problem 13 The potential is found by adding the potential due to each of the charges. Note that there are no VECTORS involved here, just numbers: P ke q 9 9 9·109 ∗8·10−6 ∗4·10−6 ∗2·10−6 + 9·10 0.03 + 9·100.0671 = 2.67 · 106 V . If we replace the bottom left charge V = r = 0.06 with −2µC, then the third term in the above sum becomes negative. This changes the answer to 2.13 · 106 V . Problem 15 Once again, add the potential due to each charge separately. Note that Q2 is negative. The potential half way between (r = 0.175m): P ke q 9 9·109 ∗5·10−9 ∗−3·10−9 V = + 9·10 0.175 = 102.86V r = 0.175 The potential energy is ∆P E = q∆V = ke q1rq2 = −3.85·10−7 , where the algebraic sign, which is negative, means that the two charges want to come closer to each other (we must intervene and push them apart to keep them separated). Problem 25 The electric field is given by E = V 20 d = 1.8·10( −3) 0 A d with 0 = = 1.1 · 104 N/C. Field lines point from positive to negative (always). The capacitace is C = 8.85 · 10−12 , which gives C = 3.74pF . Lastly the charge is given by Q = CV , which gives Q = 74.7pC on each plate. Problem 31 First, add the two capacitors in parallel, which add directly to a 6µF capacitor. Then, add that one in series with the remaining capacitor. The equivalent capacitance of the circuit is 2µF . The total charge in the circuit is then Qtot = Ceq V = 24µC. Going back one step in the reduction process (one 3µF and one 6µF in series), both of these have the same charge (24µC). Capacitors in series always have the same charge on them. Thus we have solved the 3µF capacitor:It has a charge of 24µC and a voltage of V = Q C = 8V . Going back one more step, the 6µF capacitor also has 24µC on it, which means it has a voltage of V = Q C = 4V , but remember this is a composite capacitor (it’s really made up of two capacitors). Since the two constituent 1 capacitors are in parallel, both of these capacitors have the same voltage, namely 4V. The charge on each one is then given by Q = CV . The 4µF capacitor has 16µC and the 2µF capacitor has 8µC. 2