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small identical balls with equal charges are fixed at vertices of regular 2004 gon which side is at a certain instant one of a ball is released and a sufficiently long time interval later the ball adjacent to the first released ball is fixed the kinetic energy of a released ball are found to differ by K at a sufficiently long distance from the polygon determine the charge Q of each part

small identical balls with equal charges are fixed at vertices of regular 2004 gon which side is at a certain instant one of a ball is released and a sufficiently long time interval later the ball adjacent to the first released ball is fixed the kinetic energy of a released ball are found to differ by K at a sufficiently long distance from the polygon determine the charge Q of each part

Grade:12

1 Answers

Sanghamithra
15 Points
7 months ago
(TE)1=(KE)1=k1
(TE)1=(PE)1=u
(TE)f=K1
k1=u(eq.1) 
For second ball 
 
(TE)1=u-(kq^2/q) 
(TE)f=k2=u-(kq^2/a)
K1=k2=k
u-(u-kq^2/a)=k=-1/4pieoa^2/a
Q=sqrt4pie0 (answer) 

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