solutions_28

solutions_28 - PHY2061 R D Field Chapter 28 Solutions...

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PHY2061 R. D. Field Solutions Chapter 28 Page 1 of 12 Chapter 28 Solutions Problem 1: If the potential difference between points A and B is equal to V A -V B = 3x10 4 Volts , how much work ( in milliJoules ) must be done ( against the electric force ) to move a change particle ( Q = 3 microC ) from point A to point B without changing the kinetic energy of the particle? Answer: -90 Solution: W AB = Q(V B -V A ) = (3x10 -6 C)(3x10 4 V) = 9x10 -2 J = 90 microC. We know that the work done against the electric field must be negative since the particle will fall from high potential to low potential. Problem 2: Eight identical point charges, Q , are located at the corners of a cube with sides of length L . What is the electric potential at the center of the cube? ( Take V = 0 at infinity as the reference point. ) Answer: 9.2KQ/L Solution: The potential at the center of the cube is given by V KQ d KQ L KQ L center = = = 8 16 3 9 24 . , where I used d L = 3 2 / . Problem 3: What is the electric potential energy of the configuration of eight identical point charges, Q , at the corners of the cube with sides of length L in the previous problem? Answer: 22.8KQ 2 /L Solution: The total electric potential energy is given by U QV QV Q KQ L KQ L cube i i i corner = = = + + = = 1 2 8 2 4 3 3 2 1 3 22 79 1 8 2 . , where V corner is the electric potential at the corner of the cube due to the other 7 charges.

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PHY2061 R. D. Field Solutions Chapter 28 Page 2 of 12 Problem 4: Two point charges Q 1 = Q and Q 2 = Q are on the x-axis at x = +L and x = -L and a third point charge Q 3 = Q is on the y-axis at the point y = L as shown in the Figure . What is the electric potential difference, V = V 2 -V 1 between the point P 2 at the midpoint of the line between Q 1 and Q 3 and the point P 1 at the origin? Answer: 0.5KQ/L Solution: The electric potential at P 1 is given by, L KQ V 3 1 = . At P 2 the electric potential is L KQ d KQ L KQ V + = + = 10 2 2 4 2 / 2 2 2 where 2 / 10 ) 2 ( ) 2 / 2 ( 2 2 L L L d = + = . Thus, L KQ L KQ V V 46 . 0 3 10 2 2 4 1 2 = + = . Problem 5: Two point charges Q 1 = Q and Q 2 = Q are on the x-axis at x = +L and x = - L and a third point charge Q 3 = Q is on the y-axis at the point y = L as shown in the previous Figure . What is the electric potential energy of this charge configuration? Answer: 1.9KQ 2 /L Solution: The electric potential energy of the charge configuration, U 1 , is given by, L KQ L KQ L KQ L KQ U 2 2 2 2 1 91 . 1 2 2 1 2 2 2 = + = + = . 2 x = L x = -L Q 3 =Q x-axis y = L Q 2 =Q Q 1 =Q y-axis 1 d
PHY2061 R. D. Field Solutions Chapter 28 Page 3 of 12 L P L Q Q Q Problem 6: In the problem 4 , how much work must be done ( against the electric force ) to move the charge Q 3 from the point y = L on the y-axis to the origin resulting in the configuration shown in the Figure ? Answer: 0.6KQ 2 /L Solution: The potential potential energy of this new charge configuration, U 2 , is given by, L KQ L KQ L KQ U 2 2 2 2 2 5 2 2 = + = .

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