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Question: A charge Q is divided into two parts, q and Q-q. If the coulomb repulsion between them when they are...

A charge Q is divided into two parts, q and Q-q. If the coulomb repulsion between them when they are separated is to be maximum, what should be the ratio of Q/q?
A. 22
B. 12\dfrac{1}{2}
C. 44
D. 14\dfrac{1}{4}

Explanation

Solution

Coulomb’s law is a relation about the force between two point charges. Coulomb’s law states that, “Electric force between two point charges varies inversely as the square of the distance between the charges and is directly proportional to the product of the magnitude of the two charges and acted along the line joining the two charges.”

Formula used:
Electrostatic force, Fe=kq1q2r2r^{{\mathbf{F}}_{e}}=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\hat{r}

Complete answer:
Coulomb’s law is a relation about the force between two point charges. When two linear sizes of charged bodies are separated by a distance larger as compared to their size, these can be treated as point charges.
According to Coulomb's law, “Electric force between two point charges varies inversely as the square of the distance between the charges and is directly proportional to the product of the magnitude of the two charges and acted along the line joining the two charges.”
Mathematically,
Fe=kq1q2r2r^{{\mathbf{F}}_{e}}=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\hat{r}
Where k=14πϵ0=9×109Nm2C2k=\dfrac{1}{4\pi {{\epsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}, q1{{q}_{1}} and q2{{q}_{2}} is the magnitude of charge and rr is the distance between the charges.
Since, charge Q is divided into two parts, q and Q-q.
The electrostatic force between both parts can be given as
Fe=kq(Qq)r2\left| {{\mathbf{F}}_{e}} \right|=k\dfrac{q(Q-q)}{{{r}^{2}}}
To find the charges ratio at maxima and minima, we differentiate it with respect to qqand equate to zero.
Fer=kr2(Q2q)=0\dfrac{\partial {{F}_{e}}}{\partial r}=\dfrac{k}{{{r}^{2}}}(Q-2q)=0
Q=2q\Rightarrow Q=2q
We, substitute this value in second derivative of force and get
2Feq2=2The value of charge gives us maximum force\dfrac{{{\partial }^{2}}{{F}_{e}}}{\partial {{q}^{2}}}=-2\Rightarrow \text{The value of charge gives us maximum force}
Since, Q = 2q, the ratio of Q/q is 2.

So, the correct answer is “Option A”.

Note:
The electric force between two charges varies inversely as the square of the distance between the charges and is directly proportional to the product of the magnitude of the two charges and acts along the line joining the two charges.
The force is maximum when Q = 2q