A $10 ~\mu \mathrm{C}$ charge is divided into two parts and placed at $1 \mathrm{~cm}$ distance so that the repulsive force between them is maximum. The charges of the two parts are:

The repulsive force between the two charges is given by Coulomb's law:<br/><br/> $F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2},$<br/><br/> where $F$ is the force, $q_1$ and $q_2$ are the charges, $r$ is the distance between them, and $\epsilon_0$ is the electric constant. <br/><br/> To maximize the force, we need to maximize the product $q_1q_2$. Let $q_1$ be the charge on one part and $q_2$ be the charge on the other part. Then we have $q_1+q_2=10\mu\mathrm{C}$, since the total charge is $10\mu\mathrm{C}$. <br/><br/> The distance between the two charges is $r=1\mathrm{~cm}=0.01\mathrm{~m}$. To maximize the force, we need to maximize $q_1q_2$, subject to the constraint that $q_1+q_2=10\mu\mathrm{C}$. <br/><br/> We can use the method of Lagrange multipliers to find the values of $q_1$ and $q_2$ that maximize $q_1q_2$ subject to the constraint $q_1+q_2=10\mu\mathrm{C}$. The Lagrangian is given by $\mathcal{L}=q_1q_2-\lambda(q_1+q_2-10\mu\mathrm{C}),$ where $\lambda$ is the Lagrange multiplier. <br/><br/> Taking the partial derivatives of $\mathcal{L}$ with respect to $q_1$, $q_2$, and $\lambda$, and setting them equal to zero, we get:<br/><br/> $q_2-\lambda=0$<br/><br/> $q_1-\lambda=0$<br/><br/> $q_1+q_2=10\mu\mathrm{C}$ <br/><br/> Solving for $q_1$ and $q_2$, we get $q_1=q_2=5\mu\mathrm{C}$.<br/><br/> Therefore, the charges of the two parts are both $5\mu\mathrm{C}$. <br/><br/> Therefore, to maximize the repulsive force between the two charges, we need to divide the $10\mu\mathrm{C}$ charge into two equal parts of $5\mu\mathrm{C}$ each, and place them $1\mathrm{~cm}$ apart.