Solutions to Problem set on Chapter 24

Solutions to Problem set on Chapter 24 - r= = 0.323 m....

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Unformatted text preview: r= = 0.323 m. 0.608 J r Question 5 a) From Eq. 24-35 in the textbook, we find the potential to be V =2 L / 2 + ( L2 / 4 )+ d 2 λ ln 4πε 0 d (0.06 m / 2) + (0.06 m) 2 / 4 + (0.08 m) 2 = 2(8.99 ×109 N ⋅ m 2 C2 )(3.68 ×10−12 C/m) ln 0.08 m = 2.43 ×10−2 V. b) The potential at P is V = 0 due to superposition. Question 6 We use Eq. 24-41 in the textbook. This is an ordinary derivative since the potential is a function of only one variable. d dV ˆ (−39 V/m)i. E = ˆ = x 2 )i = x)i = V/m 2 ) (0.0130 m)i = ˆ − i − (1500 ˆ ( − 3000 ˆ ( − 3000 dx dx a) Thus, the magnitude of the electric field is E = 39 V/m. b) The direction of E is −ˆ , or toward plate 1. i Question 7 The work required is: = ∆U W = Question 8 1 q1Q q2Q 1 q1Q (−q1 / 2)Q + = + = 0. 4πε 0 2d d 4πε 0 2d d Since the electric potential throughout the entire conductor is a constant, the electric potential at its center is also +400 V. ...
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