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ch25 - CHAPTER 25 Electrostatic Energy and Capacitance 1...

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CHAPTER 25 Electrostatic Energy and Capacitance 1* · Three point charges are on the x axis: q 1 at the origin, q 2 at x = 3 m, and q 3 at x = 6 m. Find the electrostatic potential energy for ( a ) q 1 = q 2 = q 3 = 2 μ C, ( b ) q 1 = q 2 = 2 μ C and q 3 = –2 μ C, and ( c ) q 1 = q 3 = 2 μ C and q 2 = –2 μ C. 1. List r 1,2 , r 1,3 , and r 2,3 ( a ), ( b ), ( c ) Use Equ. 25-1 r 1,2 = 3 m, r 1,3 = 6 m, r 2,3 = 3 m ( a ) U = 8.99 × 10 9 × 4 × 10 –12 (5/6) J = 30.0 mJ ( b ) U = –5.99 mJ ( c ) U = –18.0 mJ 2 · Point charges q 1 , q 2 , and q 3 are at the corners of an equilateral triangle of side 2.5 m. Find the electrostatic potential energy of this charge distribution if ( a ) q 1 = q 2 = q 3 = 4.2 μ C, ( b ) q 1 = q 2 = 4.2 μ C and q 3 = –4.2 μ C, ( c ) q 1 = q 2 = –4.2 μ C and q 3 = +4.2 μ C. ( a ), ( b ), ( c ) Use Equ. 25-1; note that all distances between charges are the same ( a ) U = 8.99 × 10 9 × 4.2 2 × 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 = 1/2 QV ; V = kQ / r ; U = 1/2 rV 2 / k U = 0.05 × 4 × 10 6 /8.99 × 10 9 = 22.3 μ J 4 ·· Four point charges of magnitude 2 μ C 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 r 1,2 = r 1,3 = r 2,4 = r 3,4 = 4 m; r 1,4 = r 2,3 = 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: q at (– a , + a ); 2 q at ( a , a ); –3 q at ( a , – a ); and 6 q at (– a , – a ). A fifth charge + q is 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. 1/2 mv 2 = qV (0) V (0) = ( kq/a 2 )(1 + 2 – 3 + 6) = 6 kq / a 2 v = a k/m 2 6 q 6 ·· Four identical particles each with charge Q are at the corners of a square of side L . The particles are
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