A large electroscope is made with \"leaves\" that are 78- c m -long wires with t

Question

A large electroscope is made with "leaves" that are 78-cm-long wires with tiny 27g spheres at the ends. When charged, nearly all the charge resides on the spheres.

If the wires each make a 30? angle with the vertical (see the figure(Figure 1) ), what total charge Q must have been applied to the electroscope? Ignore the mass of the wires.

A large electroscope is made with "leaves" that are 78-cm-long wires with tiny 27g spheres at the ends. When charged, nearly all the charge resides on the spheres. If the wires each make a 30? angle with the vertical (see the figure(Figure 1) ), what total charge Q must have been applied to the electroscope? Ignore the mass of the wires.

Explanation / Answer

mass of the spheres m =36 g =36*10-3 kg length of the wire L =78 cm =78*10-2 m angle ? =300 apply Newton's second law of motion, horizontal component: net force ?Fx =FTsin?-Fe =0 FTsin?-Fe =0 Fe = FTsin? ......... (1) vertical component: net force ?Fy =FT cos?-mg =0 FT = (mg)/(cos?) ....... (2) substitute the eq (2) in eq (1), we get Fe = [(mg)/(cos?)]sin? Fe = (mg) tan? ......... (3) coulomb's law states that the attractive (or) repulsive force is directly proportional to product of two charges and inversely proportional to square of the distance between two charges. i.e., Fe =(1/4??0)q1q2/r2 from given problem we have chrge =Q/2 so, electrostatic force Fe =(1/4??0)(Q/2)(Q/2)/r2 let r =L Fe =(1/4??0)(Q/2)2/L2 ........... (4) compare eq (3) and (4), we get (1/4??0)(Q/2)2/L2 = (mg) tan? where,k = (1/4??0) =8.99*109 N.m2/C2 (1/4??0)(Q/2)2/L2 = (mg) tan? by solving we get charge Q =2L?(mg tan?)/(k) ............ (5) substitute the given data in eq (5), we get Q = 7.42*10-6 C = 7.42 ?C apply Newton's second law of motion, horizontal component: net force ?Fx =FTsin?-Fe =0 FTsin?-Fe =0 Fe = FTsin? ......... (1) vertical component: net force ?Fy =FT cos?-mg =0 FT = (mg)/(cos?) ....... (2) substitute the eq (2) in eq (1), we get Fe = [(mg)/(cos?)]sin? Fe = (mg) tan? ......... (3) coulomb's law states that the attractive (or) repulsive force is directly proportional to product of two charges and inversely proportional to square of the distance between two charges. i.e., Fe =(1/4??0)q1q2/r2 from given problem we have chrge =Q/2 so, electrostatic force Fe =(1/4??0)(Q/2)(Q/2)/r2 let r =L Fe =(1/4??0)(Q/2)2/L2 ........... (4) compare eq (3) and (4), we get (1/4??0)(Q/2)2/L2 = (mg) tan? where,k = (1/4??0) =8.99*109 N.m2/C2 (1/4??0)(Q/2)2/L2 = (mg) tan? by solving we get charge Q =2L?(mg tan?)/(k) ............ (5) substitute the given data in eq (5), we get Q = 7.42*10-6 C = 7.42 ?C (1/4??0)(Q/2)2/L2 = (mg) tan? by solving we get charge Q =2L?(mg tan?)/(k) ............ (5) substitute the given data in eq (5), we get Q = 7.42*10-6 C = 7.42 ?C

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