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1:00 AM- Due Date: C9%) Problem, 9: A positively charged particle Q,--25 nC is held fixed at the origin. A second charge o, of mass . 6.5 μ¡s floating a distance d= 25 cm above charge , The net force on Q2 is equal to zero. You may assume this system is close to the surface of Q2 the Earth Ctheexpertta.com to any sabt ons ณ ttg website = strictly forbidden. Doing so may renz Atepts renaunng
12-01.00 AM-Due Date: 1/2 00 AM (9%) Problem 9: A positively charged particle 2,-+25 nCis held fixed at the origin A second charge Q2 of mass m-6.5 ug is floating a distance d- 25 cm above charge Q1 The net force on 02 is equal to zero. You may assume this system is close to the surface of the Earth alnajrani, Q2 tracking id: 7072-47 SE-46-8475-19549. In accordance with Expert TAs Tenss of to any solutions sharing website is strictly forbidden Doing so may resvalt in termination of your Expert TA Account 25% Part (a) which of the following represents the fr is the sign of the charge on 2? Grade Sammary Q, is unchargedNegativePositive Potential 100% Attempts remaining 2 50% per attesp) Hint I give up 呈▲ 25% Part (d) Calculate the magnitude of Q2 in units of
t I Help 9 12:01:00 AM--Due Date: 1/2 О%) Problem 9: A positively charged particle ),-+25 nC is held fixed at the origin A second charge o, of mass m-6.5 μg ís floating a distance d= 25 crn above charge Q1 The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth 02 呈血25% Part (b) What is the sign of the charge on Q, of charge Q, in terms of the given variables Grade Summary Potential Sabmissions % per attempt) 1 2 3 Hints: 2% deduction per hntHintstem of Q2 in units of LLC
#01 (9%) Problem 9: A second charge Q2 of mass m-6.5 ug is floating a distance d - 25 em above chargeQ The net force on Q2 is equal to zero. You may assume this system is close to the surface of charged particle Q +25 nC is held fixed at the origin A 25% Part (b) What is the sign of the charge on of 0, in units of nanocoulombs Potential 100% attempt) cotan0 asinOacos0 1 2 3
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Answer #1

The electric force is repulsive and gravitational force is attractive these two are balanced so the correct free body diagram is Like charges repel each other unlike charges attract each other so the sign of charges Q1 and Q2 is same That is sign of charge Q2 is positive The balancing equation is mgd The charge 1S mgd _( 6.5x10-6)(9.8) (0.25) 9x109)(25x109) -17.69x10C 17.69 nC

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