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A force F acts between two equal and opposite charges which are placed at a certain distances apart .If 25% of one charge is transferred to the other then what is the force between them?
Asked by Vidushi412 | 28 Mar, 2019, 11:25: AM
answered-by-expert Expert Answer
Given F is the force between the equal and opposite charges. 
 
25 % of the charge is transferred to another. 
 
Thus, + q becomes,
 
3 over 4 q
 
Thus, - q becomes,
negative q space plus space 1 fourth q space... left parenthesis space a s space 25 space percent sign space o f space c h a r g e space i s space t r a n s f e r r e d space f r o m space plus q space t o space minus q right parenthesis 
 
= fraction numerator negative 3 over denominator 4 end fraction q
By Coulomb's law, 
 
F apostrophe space equals space fraction numerator k q subscript 1 q subscript 2 over denominator r squared end fraction equals fraction numerator k begin display style 3 over 4 end style q open parentheses begin display style fraction numerator negative 3 over denominator 4 end fraction end style q close parentheses over denominator r squared end fraction F apostrophe space equals negative k space q squared fraction numerator begin display style bevelled 9 over 16 end style over denominator r squared end fraction space equals space 9 over 16 open parentheses negative fraction numerator k q squared over denominator r squared end fraction close parentheses space equals space 9 over 16 F
 
Thus, the force between the charges become 9 over 16of the original value. 
Answered by Shiwani Sawant | 28 Mar, 2019, 01:04: PM
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