Consider the two charges: q1 and q2 that are lined up and separated by 4.5 cm an

Question

Consider the two charges: q1 and q2 that are lined up and separated by 4.5 cm and fixed in place. Charge Q= 1.62*10^-6 C is placed as shown ( halfway between the two points such that they form a triangle with the other two sides equal to 3 cm). If q accelerates upward from rest with a value of 299 m/s^2 and measures 5 g, determine the value of charges q1 and q2 and their signs.

What I know: F=ma ==> Upward force = 14.95N

Since Q is positive q1 has to be negative and q2 has to be positive

Q1 has to be larger to push Q away

. q1

3cm/ |

/ / |

| / | 4.5 cm

Q . |

|

3cm |

|

.q2

Explanation / Answer

pick a value of +1µC for the third charge set the origin at q1 14 cm = 0.14 m Coulomb's law, force of attraction/repulsion F = kQ1Q2/r² Q1 and Q2 are the charges in coulombs r is separation in meters k = 8.99e9 Nm²/C² (if the charges are in µC, you can use k = 8.99e–3) Force between q1 and q3 is F1 = k(3)(1) / (x)² = 3k/x² Fore is to the right, call this positive Force between q2 and q3 is F1 = k(30)(1) / (0.14–x)² = 30k/(x²–0.28x+0.0196) Force is to the right, also positive Total force, F is F = 3k/x² + 30k/(x²–0.28x+0.0196) simplify and take the derivative, and set it equal to 0, and solve for x ,

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