Dear Student, let the charge travel d distnace after it stop potential energy of the charge Q at this point P=-2k259+d210-12 due to two chrges potential energy at C P'=-2k25310-12initial PE - final PE=P'-P=-2k25310-12+2k259+d210-12-2k25310-12+2k259+d210-12=-0 060 459+d2=0 099+d2=25d=4m Regards

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Dear Student,

l e t t h e c h a r g e t r a v e l d d i s t n a c e a f t e r i t s t o p . p o t e n t i a l e n e r g y o f t h e c h a r g e Q a t t h i s p o i n t . P = \frac{- 2 (k 25)}{\sqrt{9 + d^{2}}} 10^{- 12} d u e t o t w o c h r g e s . p o t e n t i a l e n e r g y a t C P' = \frac{- 2 (k 25)}{3} 10^{- 12} i n i t i a l P E - f i n a l P E = P' - P = \frac{- 2 (k 25)}{3} 10^{- 12} + \frac{2 (k 25)}{\sqrt{9 + d^{2}}} 10^{- 12} \frac{- 2 (k 25)}{3} 10^{- 12} + \frac{2 (k 25)}{\sqrt{9 + d^{2}}} 10^{- 12} = - 0.06 \frac{0.45}{\sqrt{9 + d^{2}}} = 0.09 9 + d^{2} = 25 d = 4 m R e g a r d s

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