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Why dF/dq is equal to zero for the force to be maximum in the problem below: A charge Q is divided into two parts of qand Q-q if the coulomb repulsion between them when they are seprated is to be maximum the ratio of Q/q should be

Why dF/dq is equal to zero for the force to be maximum in the problem below:



A charge Q is divided into two parts of qand Q-q if the coulomb repulsion between them when they are seprated is to be maximum the ratio of Q/q should be


Grade:12

4 Answers

Mujahid Ahmed
42 Points
11 years ago

It''s a basic result. For a point to be local maximum, the tangent at that point must be zero. Tangent is given by dF/dQ. 

 

 

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saurabh gupta
14 Points
11 years ago

R/21/2

FITJEE
43 Points
11 years ago

IT IS BECAUZ THE AREA UNDER THE CURVE OF FORCE VS CHARGE GRAPH GIVE POTENTIAL AND DF/DQ GIVE TANGENT AT A PARTICULAR POINT AND IF SLOPE OF TANGENT AT A POINT IS 0 OR MINIMUM THEN FORCE WILL BE MAXXXXXXXXX..........

Abhishek Sharma
13 Points
5 years ago
In application of derivatives we have studied that .  If we have to find maximum value ex :
                       y = x^2+x+1
                      dy/dx = 2x+1
                and dy/dx = 0
                     x = ½
         And dy/dx = tan z ,where z is an angle formed with x axis.    
          At any point the graph of this equation it forms straight line  which shows that
               dy/dx = 0
                  

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