ResponseDetails: A charged cork ball of mass 1.0 gram is suspended on a stringin

Question

ResponseDetails: A charged cork ball of mass 1.0 gram is suspended on a stringin the presence of a uniform electric field. The electric field isoriented 45 degrees to the horizontal and has magnitude of 5x10^5N/C. Find the angle at whoich the charged ball will be inequilibrium and find the tension in the string. Please answer with all givens provided. This is not an easystraight forward problem and combines many concepts. NOT ATEXT PROBLEM Moderators answer was incorrect. No credt willbe given unless all values are included in an accuratesolution A charged cork ball of mass 1.0 gram is suspended on a stringin the presence of a uniform electric field. The electric field isoriented 45 degrees to the horizontal and has magnitude of 5x10^5N/C. Find the angle at whoich the charged ball will be inequilibrium and find the tension in the string. Please answer with all givens provided. This is not an easystraight forward problem and combines many concepts. NOT ATEXT PROBLEM Moderators answer was incorrect. No credt willbe given unless all values are included in an accuratesolution

Explanation / Answer

E field UP or DOWN? maybe we can figure it out. . If the E field was up, the vertical electric force wouldbe   qE sin45 = 1 x 10-9 * 5 x105 * sin45 = 3.54 x 10-4 N .    The downward force of gravity is  mg = 0.001 * 9.8 =   9.8 x10-3 . Hmmm... well, what I was thinking was that if the verticalforce from the E field was greater than mg, then the E field wouldhave to be downward. But since mg is greater, the E field couldeither be above or below the horizontal and you can get a validanswer in either case. . Yep... your prof is a bit absent-minded. . Hmmm... I'll assume the E field is above the horizontal. . Then...                 horizontal forces:    qE cos45 - Tcos = 0 .                    vertical forces:    qE sin45 + Tsin - mg = 0 .              Now you have two equations and two unknowns... .     T sin   = mg - qE sin45 .     T cos   =   qEcos45 . Divide the equations:     tan = ( mg - qE sin45) / qEcos45   = .                         =   mg/qE cos45   - tan45  = .                 = 0.001 * 9.8 / 1.0 x 10-9 * 5.0 x105 * cos45      - 1   = .             =   27.72 - 1   =  26.72       .      =   87.93 degrees above the horizontal .       T cos87.93 = qE cos45 .      T = 1 x 10-9 *5.0 x 105 cos45 / cos87.93   =   9.453 x 10-3Newtons

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