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Answers to Chapter 16 Problems

 

16.5.1   (a)  0.414 m     (b) 1.88 ×105 V m-1  at an angle of 4.89° to the line between the charges and above this line

16.5.2   +4.0 × 10–12 C,  q2 = –2.0 × 10–12 C   or   â€“4.0 × 10–12 C,  q2 = +2.0 × 10–12 C

16.5.3   (a)  7.5 × 10–3 C m–2    (b)  8.4 × 108 V m–1

16.5.4   375 V m−1

16.6.1   (a)  3.1× 10−6 V m−1   (b)  1.4 × 10−6 V m−1

16.8.1   T = \( {2\pi \over e }\sqrt{4 \pi \epsilon_0 R^3 m} \);    893 Hz

16.8.2   (i)  2.70 × 104 V m−1      (ii)  3.33 × 103 V m−1

16.8.3   1.17 × 1016 electrons;    8.17 × 10–9

16.8.4    \( {e^2\over 4 \pi \epsilon_0 a^2} \Bigl[ {3 \over 2 \sqrt{2}} \pmb i + \Bigl(2 + {3 \over 2 \sqrt{2}} \pmb j \Bigr) \Bigr]\);    2.4×10–11 N  at 19° with the horizontal

16.8.5     (a)  4 × 10−40    (b)  8.2 × 10−8 N

16.8.6 

16.9.1   \( E_1 = {Q\over 4 \pi \epsilon_1 r^2} \), when \( {a < r < R } \);      \(  E_2 = {Q\over 4 \pi \epsilon_2 r^2} \), when \( {R < r < b } \);      \( {Q\over 4 \pi } \Bigl[ \Bigl( {a-R \over \epsilon_1 aR} \Bigr ) + \Bigl( {R-b \over \epsilon_2 bR} \Bigr) \Bigr]\)

16.10.1  (a) 0.56 × 10–2  N in the direction A → C    (b) 4490 V m–1 in the direction A → C    (c) 1.92 × 104 V 

16.10.2   VP = \( {6\sqrt{2} e \over 4 \pi \epsilon_0 a} \);    2.2 V

16.10.3   VA = 3 × 10–8 V;    VB = 0

16.12.1   V = \( Q \over {4 \pi \epsilon_0 {(x^2 + R^2)}^{3/2}} \)

16.13.1   2.24 × 10–12  J;    14 Mev;     2.6 × 107 m s–1 

16.13.2   1.3 × 108 m s−1 

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