Question

Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases $4.5$ times in comparison with the initial value. The ratio of the initial charges of the balls is ....... (A) $4:1$ (B) $6:1$ (C) $3:1$ (D) $2:1$

Short answer

The ratio of the initial charges of the balls is $\boxed{{\displaystyle \text{(C)}3:1}}$.

Step-by-step solution

Write down the Coulomb's law formula

Coulomb's law states that the force between two point charges is given by: $$F=k\frac{q_1{q}_2}{r^2}$$ where, $F$ is the force between charges, $q_1$ and $q_2$ are the magnitudes of charges, $r$ is the separation between charges, and $k$ is Coulomb's constant.

Set up two equations for the two cases

Let the initial charges be $q_1$ and $q_2$. The initial force and separation can be represented as $F_1$ and $r_1$. Using the Coulomb's law, $$F_1=k\frac{q_1{q}_2}{(r_1{)}^2}$$ When the balls are brought in contact, the charges become $\frac{q_1+q_2}{2}$ for both balls. Then, they are moved apart to a distance of $\frac{r_1}{2}$. The force between them increases 4.5 times to become $4.5F_1$. $$4.5F_1=k\frac{{\left(\frac{q_1+q_2}{2}\right)}^2}{{\left(\frac{r_1}{2}\right)}^2}$$

Manipulate the equations to eliminate variables

From the first equation, we can write: $$k\frac{q_1{q}_2}{(r_1{)}^2}=F_1$$ Now, substituting this expression for $F_1$ in the second equation: $$4.5k\frac{q_1{q}_2}{(r_1{)}^2}=k\frac{{\left(\frac{q_1+q_2}{2}\right)}^2}{{\left(\frac{r_1}{2}\right)}^2}$$ Cancelling out $k$ and $(r_1{)}^2$ from both sides, and simplifying, we get: $$4.5(q_1{q}_2)={\left(\frac{q_1+q_2}{2}\right)}^2$$ Expanding and rearranging: $$\frac{q_1^2}{2}+\frac{q_2^2}{2}=3.5q_1{q}_2$$

Solve for the ratio of the charges, $q_1$ and $q_2$

Divide the equation by $q_1{q}_2$: $$\frac{q_1}{2q_2}+\frac{q_2}{2q_1}=3.5$$ Let $\frac{q_1}{q_2}=x$, then the equation becomes a quadratic equation in $x$: $$\frac{x}{2}+\frac{1}{2x}=3.5$$ Solving the quadratic equation, we get $x=\frac{q_1}{q_2}=3$, which means the ratio of the initial charges is $3:1$. Therefore, the answer is $\boxed{{\displaystyle \text{(C)}3:1}}$.

Key concepts

Electrostatic Force

Electrostatic force is a fundamental concept in physics, describing the force acting between electrically charged objects. It strictly follows Coulomb's Law, which expresses how the force depends on the magnitude of the charges and the distance between them. The governing formula is:

• $F=k\frac{q_1{q}_2}{r^2}$

Where:

• $F$ is the electrostatic force.
• $q_1$ and $q_2$ represent the amount of the charges.
• $r$ is the separation distance between the charges.
• $k$ is Coulomb’s constant, approximately equal to $8.99\times {10}^9{\text{ N m}}^2/{\text{C}}^2$.

It's important to note:

• Like charges repel each other, while opposite charges attract.
• The force becomes stronger as the charges become larger or the separation distance becomes smaller.

Understanding electrostatic force is key to solving problems involving electrical interactions in physics.

Charge Distribution

Charge distribution refers to how electric charge is arranged within a system of objects. In scenarios where objects are brought into contact, like in our exercise, the distribution of independent charges can change dramatically.

• When two identically charged objects are touched together, they share their total charge equally.
• This is because charge migrates across a conductive path to minimize potential energy differences.

Upon separating them again, each object ends up with an average of the initial total charge. For instance, if two balls with charges $q_1$ and $q_2$ touch, after separation, they each carry $\frac{q_1+q_2}{2}$ charge.

Understanding charge distribution is crucial when calculating electrostatic forces post-contact, such as when their separation changes or their post-contact charge impacts future interactions. It allows us to account for changes accurately in physical systems.

Quadratic Equation

Quadratic equations come into play directly when trying to find the ratio of charges in problems involving electrostatics. They are of the form:

• $a{x}^2+bx+c=0$

In our exercise, by letting $x=\frac{q_1}{q_2}$, we could transform the relationship into a quadratic equation:

• $\frac{x}{2}+\frac{1}{2x}=3.5$ simplifies to $2x^2-7x+2=0$

Solving gives us possible values for $x$, which represents the ratio of the initial charges. The quadratic formula used frequently is:

• $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, the solution results in $x=3$, giving a ratio of $3:1$. Quadratic equations offer a way to handle squared terms effectively, especially in distribuitive or rearranged scenarios stemming from physical laws.