Particle 3 be placed to minimize the magnitude of

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particle 3 be placed to minimize the magnitude of that force? Answer The coordinate system is already defined for us. 8.00cm q 1 =+e q 2 =-27e +x x=0
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Superposition Answer Next, let’s list the quantities that we know: q 1 = +e q 2 = -27e q 3 = +4e r 23 = ? r 13 = ? Let’s alter the diagram of the situation by placing q 3 somewhere; call it a distance x from q 1 . q 1 =+e q 2 =-27e +x x=0 x=8 x This means that we now have a value for r 13 (x) and a value for r 23 (8-x) What will the direction of q1 on q3 be? What will the direction of q2 on q3 be? To the right. + To the right.
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Superposition Answer Since both forces point in the same direction, we just need to add the absolute values of the forces involved. Putting in the values gives us: F net = F 13 + F 23 F net = k q 1 q 3 x 2 + k q 2 q 3 8 cm ! x ( ) 2 F net = 4 ke 2 1 x 2 + 27 8 cm ! x ( ) 2 " # $ $ % & ' ' In order to minimize this force, we need to take a derivative with respect to x and set it equal to zero. dF net dx = 4 ke 2 ! 2 x 3 + ! 1 ! 2 ( ) 27 ( ) 8 cm ! x ( ) 3 " # $ $ % & ' ' = 0
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Superposition Answer The only part that can go to zero is inside the parenthesis, so we can set that to zero: Eliminating the denominators yields: Taking a cube root of both sides gives us: ! 2 x 3 + 54 8 cm ! x ( ) 3 " # $ $ % & ' ' = 0 2 x 3 = 54 8 cm ! x ( ) 3 8 cm ! x ( ) 3 = 54 x 3 2 = 27 x 3 8 cm ! x ( ) = 3 x 8 cm = 4 x x = 2.00 cm This means particle 3 is placed 2.00 cm from particle 1 to minimize the force on it.
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For Next Time (FNT) Start reading Chapter 22 Start working on the Chapter 21
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