CHAPTER25Electrostatic Energy and Capacitance1* ·Three point charges are on the xaxis: q1at the origin, q2at x= 3 m, and q3at x= 6 m. Find the electrostatic potential energy for (a) q1= q2= q3= 2 μC, (b) q1= q2= 2 μC and q3= –2 μC, and (c) q1= q3= 2 μC and q2= –2 μC. 1. List r1,2, r1,3, and r2,3(a), (b), (c) Use Equ. 25-1 r1,2= 3 m, r1,3= 6 m, r2,3= 3 m (a) U= 8.99×109×4×10–12(5/6) J = 30.0 mJ (b) U= –5.99 mJ (c) U= –18.0 mJ 2 ·Point charges q1, q2, and q3are at the corners of an equilateral triangle of side 2.5 m. Find the electrostatic potential energy of this charge distribution if (a) q1= q2= q3= 4.2 μC, (b) q1= q2= 4.2 μC and q3= –4.2 μC, (c) q1= q2= –4.2 μC and q3= +4.2 μC. (a), (b), (c) Use Equ. 25-1; note that all distances between charges are the same (a)U= 8.99×109×4.22×10–12(3/2.5) J = 0.19 J (b)U= 0.0634 J; (c) U= –0.0634 J 3· What is the electrostatic potential energy of an isolated spherical conductor of radius 10 cm that is charged to 2 kV? U= 1/2QV; V= kQ/r; U= 1/2rV2/kU= 0.05×4×106/8.99×109= 22.3 μJ 4 ··Four point charges of magnitude 2 μC are at the corners of a square of side 4 m. Find the electrostatic potential energy if (a) all the charges are negative, (b) three of the charges are positive and one is negative, and (c) two are positive and two are negative. 1. Determine the distances between charges (a), (b), (c) Use Equ. 25-1 r1,2= r1,3= r2,4= r3,4= 4 m; r1,4= r2,3= 5.657 m (a)U= 48.7 mJ; (b) U= 0; (c) U= –12.7 mJ or –23.2 mJ, depending on the configuration 5* ··Four charges are at the corners of a square centered at the origin as follows: qat (–a, +a); 2qat (a, a); –3qat (a, –a); and 6qat (–a, –a). A fifth charge +qis placed at the origin and released from rest. Find its speed when it is a great distance from the origin. 1. Find V(0) 2. 1/2mv2= qV(0) V(0) = (kq/a2)(1 + 2 – 3 + 6) = 6kq/a2v= ak/m26q6 ··Four identical particles each with charge Qare at the corners of a square of side L. The particles are
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