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, the ratio of q/Q should be?
Understand the Problem
The question is about dividing a charge Q into two parts and determining the ratio of one part to the total charge that maximizes the Coulomb repulsion between the two parts. To solve this, we will apply principles of electrostatics.
Answer
The ratio is $\frac{1}{2}$.
Answer for screen readers
The ratio of $\frac{q}{Q}$ that maximizes the Coulomb repulsion is $\frac{1}{2}$.
Steps to Solve
- Define the charges and forces involved
Let the total charge be $Q$, divided into two parts: $q$ and $Q - q$. The Coulomb force (repulsion) between the two charges is given by the formula:
$$ F = k \frac{q (Q - q)}{r^2} $$
where $k$ is Coulomb's constant, and $r$ is the distance between the charges.
- Express the force in terms of q
Rewrite the expression for the force:
$$ F = k \frac{q(Q - q)}{r^2} $$
- Differentiate the force with respect to q
To find the value of $q$ that maximizes the force, we take the derivative of $F$ with respect to $q$ and set it to zero:
$$ \frac{dF}{dq} = k \frac{(Q - 2q)}{r^2} = 0 $$
- Solve for q
Set the derivative to zero and solve for $q$:
$$ Q - 2q = 0 $$
This leads to
$$ q = \frac{Q}{2} $$
- Find the ratio of $\frac{q}{Q}$
Now calculate the ratio of $q$ to the total charge $Q$:
$$ \frac{q}{Q} = \frac{\frac{Q}{2}}{Q} = \frac{1}{2} $$
The ratio of $\frac{q}{Q}$ that maximizes the Coulomb repulsion is $\frac{1}{2}$.
More Information
When the charge is equally split into two parts, it maximizes the electrostatic repulsion between them due to the squared dependence of the force on the distance and the product of the two charges.
Tips
- Failing to set the derivative to zero or miscalculating the derivative can lead to incorrect values for $q$.
- Confusing the parts of the charge can result in mixing up the values and the ratio.
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