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Unformatted text preview: CHAPTER 26 ELECTROSTATIC ENERGY AND CAPACITORS Section 26-1: Energy of a Charge Distribution Problem 1. Three point charges, each of + q , are moved from infinity to the vertices of an equilateral triangle of side . How much work is required? Solution The sentence preceding Example 26-1 allows us to rewrite Equation 26-1 (for the electrostatic energy of a distribution of point charges) as W kq q r i j ij = pairs = . For three equal charges (three different pairs) at the corners of an equilateral triangle ( r ij = for each pair) W kq = 3 2 = . Problem 2. Repeat the preceding problem for the case of two charges + q and one - q . Solution When two pairs have opposite charges ( ), q q q i j = - 2 and one pair has equal charges ( ), q q q i j = 2 the electrostatic energy is W k q q q kq =-- + = - ( ) . 2 2 2 2 = = Problem 3. Four 50- C charges are brought from far apart onto a line where they are spaced at 2.0-cm intervals. How much work does it take to assemble this charge distribution? Solution Number the charges q i i = = 50 1 2 3 4 C , , , , , as they are spaced along the line at a = 2 cm intervals. There are six pairs, so W kq q r k q q a q q a q q a q q a q q a q q a kq a i j ij = = + + + + + = + + + + + = pairs = = = = = = = = ( ) ( )( ) 1 2 1 3 1 4 2 3 2 4 3 4 2 1 2 1 3 1 2 2 3 2 1 1 1 13 3 13 9 10 50 3 2 4 88 2 9 2 kq a = = = = ( / )( ) ( ) . . m F C cm kJ (See solution to Problem 1.) Problem 3 Solution. Problem 4. Repeat Example 26-1 for the case when the negative charge is - q rather than - q = 2. Solution If the negative charge in Example 26-1 is - q W , 2 and W 3 are unchanged, but W k q a q a q a 4 2 2 2 2 =--- ( ). = = = Therefore, W W W W = + + = 2 3 4 0. (In this case, the work needed to assemble the positive charges equals the energy gained adding the negative charge.) Problem 5. Suppose two of the charges in Problem 1 are held in place, while the third is allowed to move freely. If this third charge has mass m , what will be its speed when its far from the other two charges? 2 CHAPTER 26 Solution With one charge removed to infinity, the potential energy is reduced to that of just one pair of charges, W kq f = 2 = . The initial potential energy was W kq i = 3 2 = (see Problem 1), so the kinetic energy of the charge at infinity (from the conservation of energy) is K W W kq i f =- = 2 2 = . Thus, v = = = = 2 K m q m . Problem 6. To a very crude approximation, a water molecule consists of a negatively charged oxygen atom and two bare protons, as shown in Fig. 26-25. Calculate the electrostatic energy of this configuration, which is therefore the magnitude of the energy released in forming this molecule from widely separated atoms. Your answer is an overestimate because electrons are actually shared among the three atoms, spending more time near the oxygen....
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