Question : A total charge Q is broken in two parts Q1 and Q2 and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when
a)Q2=Q/R, Q1=Q-Q/R
b)Q2=Q/4, Q1=Q-2Q/3
c)Q2=Q/4, Q1=3Q/4
d)Q1=Q/2, Q2=Q/2
Doubt by Rimjhim
Solution :
Q=Q1+Q2
So, Q2=Q-Q1
According to Coulomb's law
F=kQ1Q2/R2
F=kQ1(Q1)(Q-Q1)/R2
F=kQ1(Q1Q-Q12)/R2
Applying the concept of maxima and minima from Application of Derivatives.
For F to be maximum, dF/dQ1=0
Differentiating the above equation w.r.t. Q1
dF/dQ1=(k/R2)(Q-2Q1)
0=Q-2Q1
2Q1=Q
Q1=Q/2
Therefore, Q1=Q2=Q/2
This is the required condition for maximum repulsion.
Similar Question :
A charge Q is divided into two parts of q and Q-q. If the coulomb repulsion between them when they are separated is to be maximum, then ratio of Q/q should be
a) 2:1
b) 1/2
c) 4:1
d) 1/4
Doubt by Tushar
Correct Answer : a) 2:1
