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Question: A total charge of 20 \(\mu C\) is divided into two parts and placed at some distance apart. If the c...

A total charge of 20 μC\mu C is divided into two parts and placed at some distance apart. If the charges experience maximum coulombic repulsion, the charges should be:
(A) 5μC,15μC5\mu C,15\mu C
(B) 10μC,10μC10\mu C,10\mu C
(C) 12μC,8μC12\mu C,8\mu C
(D) 403μC,203μC\dfrac{40}{3}\mu C,\dfrac{20}{3}\mu C

Explanation

Solution

Consider the divided charges as q and (20q)(20-q) places at some distance apart. To calculate the maximum coulombic repulsion between the charges, we equate the differentiation of Coulomb force with respect to charge to zero and then calculate the value of charge q.

Complete step by step answer:
According to the question, a total charge of 20 μC\mu C is divided into two parts and placed at some distance apart, say R.

Let us consider the two charges as q and (20q)(20-q)as shown in the figure below.


The two charges are of the same magnitude so they repel each other and the formula for coulomb repulsion is given by,
F=14π0q1q2R2F=\dfrac{1}{4\pi {{\in }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{R}^{2}}} ….(i)
Here,
q1=q{{q}_{1}}=q and q2=20q{{q}_{2}}=20-q

So, putting the above values in equation (i) we get,
F=14π0q(20q)R2F=\dfrac{1}{4\pi {{\in }_{0}}}\dfrac{q(20-q)}{{{R}^{2}}} ….(ii)

Now, for the maximum coulombic repulsion between the charges q and (20q)(20-q), we differentiate the Coulomb force F with respect to charge q, we get,
dFdq=14π0R2ddq[q(20q)]\dfrac{dF}{dq}=\dfrac{1}{4\pi {{\in }_{0}}{{R}^{2}}}\dfrac{d}{dq}[q(20-q)]
dFdq=14π0R2ddq[20qq2)]\Rightarrow \dfrac{dF}{dq}=\dfrac{1}{4\pi {{\in }_{0}}{{R}^{2}}}\dfrac{d}{dq}[20q-{{q}^{2}})]
dFdq=202q4π0R2\Rightarrow \dfrac{dF}{dq}=\dfrac{20-2q}{4\pi {{\in }_{0}}{{R}^{2}}} ….(iii)

Now for calculating the value of q we equate the maximum coulombic repulsion between the charges q and (20q)(20-q) i.e. equation (iii) to zero.
dFdq=0\dfrac{dF}{dq}=0
202q4π0R2=0\Rightarrow \dfrac{20-2q}{4\pi {{\in }_{0}}{{R}^{2}}}=0
202q=0\Rightarrow 20-2q=0
q=10μC\Rightarrow q=10\mu C
Also,
20q=(2010)μC=10μC\Rightarrow 20-q=(20-10)\mu C=10\mu C

Hence, the two charges experience maximum coulombic repulsion when they are divided in equal parts i.e. 10μC\mu C and 10μC\mu C.

Therefore, the correct answer is option (B).

Additional Information:
The formula for coulomb force is given by,
F=kq1q2R2F=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{R}^{2}}}
Where, k=14π0k=\dfrac{1}{4\pi {{\in }_{0}}}is the Coulomb constant, also called the electric force constant in electrostatics is,
k=8.9875517923×109kgm3s2C2k=8.9875517923\times {{10}^{9}}kg\cdot {{m}^{3}}\cdot {{s}^{-2}}\cdot {{C}^{-2}}
And
0=8.8541878128×1012Fm1{{\in }_{0}}=8.8541878128\times {{10}^{-12}}F\cdot {{m}^{-1}} is the permittivity of free space or vacuum.

Note:
For solving such questions students should memorize the formula for Coulomb force between two charges placed some distance apart. Also the concept of calculating maximum force experienced by equating the derivative to zero should be remembered.