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Chapter 11: Problem 1524

A charge \(\mathrm{Q}\) is divided into two parts and then they are placed at a fixed distance. The force between the two charges is always maximum when the charges are \(\ldots \ldots\) (A) \((Q / 3),(Q / 3)\) (B) \((\mathrm{Q} / 2),(\mathrm{Q} / 2)\) (C) \((Q / 4),(3 Q / 4)\) (D) \((Q / 5),(4 Q / 5)\)

Short Answer

Expert verified
The optimal way to divide the charge is to have both charges equal. Therefore, the charges should be \(\frac{Q}{2}\) and \(\frac{Q}{2}\), as in option (B).

Step by step solution

01

Understand the problem

We are given a charge Q which is divided into two parts say q1 and q2 such that q1+q2=Q. We're tasked to find in what ratio the charges should be divided so as to maximize the force between them.
02

Coulomb's Law

Coulomb's Law states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it can be written as: \[F = k\frac{q1 \cdot q2}{r^2}\] where \(F\) is the force between the charges, \(k\) is Coulomb's constant, \(q1\) and \(q2\) are the charges, and \(r\) is the distance between the charges.
03

Express q2 in terms of q1

Since q1 + q2 = Q, we can express q2 in terms of q1 as follows: \[q2 = Q - q1\]
04

Substitute q2 in the Coulomb's Law equation

We can now replace q2 in the Coulomb's Law equation with the expression we found above: \[F = k\frac{q1 \cdot (Q - q1)}{r^2}\]
05

Maximize the Force

For the force to be maximum, the product q1 * q2 should be maximum. We can rewrite the equation for \(F\) and eliminate the constant values: \[F' = q1 \cdot (Q - q1)\]
06

Find the first derivative of F'

To find the maximum value, we need to find the first derivative of F' with respect to q1 and set it to zero: \[\frac{dF'}{dq1} = (Q - q1) - q1 = Q - 2q1\]
07

Set the derivative to zero and solve for q1

Set the first derivative equal to zero and solve for q1: \[Q - 2q1 = 0\] \[q1 = \frac{Q}{2}\]
08

Substitute q1 back into q2 expression

Now, we can substitute the value of q1 back into the expression for q2: \[q2 = Q - \frac{Q}{2} = \frac{Q}{2}\] Therefore the answer is (B) \((\mathrm{Q} / 2),(\mathrm{Q} / 2)\).

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Most popular questions from this chapter

Two positive point charges of \(12 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed \(10 \mathrm{~cm}\) apart in air. The work done to bring them $4 \mathrm{~cm}$ closer is (A) Zero (B) \(4.8 \mathrm{~J}\) (C) \(3.5 \mathrm{~J}\) (D) \(-5.8 \mathrm{~J}\)

Two parallel plate air capacitors have their plate areas 100 and $500 \mathrm{~cm}^{2}$ respectively. If they have the same charge and potential and the distance between the plates of the first capacitor is \(0.5 \mathrm{~mm}\), what is the distance between the plates of the second capacitor ? (A) \(0.25 \mathrm{~cm}\) (B) \(0.50 \mathrm{~cm}\) (C) \(0.75 \mathrm{~cm}\) (D) \(1 \mathrm{~cm}\)

At what angle \(\theta\) a point \(P\) must be located from dipole axis so that the electric field intensity at the point is perpendicular to the dipole axis? (A) \(\tan ^{-1}(1 / \sqrt{2})\) (B) \(\tan ^{-1}(1 / 2)\) (C) \(\tan ^{-1}(2)\) (C) \(\tan ^{-1}(\sqrt{2})\)

The circular plates \(\mathrm{A}\) and \(\mathrm{B}\) of a parallel plate air capacitor have a diameter of \(0.1 \mathrm{~m}\) and are $2 \times 10^{-3} \mathrm{~m}\( apart. The plates \)\mathrm{C}\( and \)\mathrm{D}$ of a similar capacitor have a diameter of \(0.1 \mathrm{~m}\) and are $3 \times 10^{-3} \mathrm{~m}\( apart. Plate \)\mathrm{A}\( is earthed. Plates \)\mathrm{B}$ and \(\mathrm{D}\) are connected together. Plate \(\mathrm{C}\) is connected to the positive pole of a \(120 \mathrm{~V}\) battery whose negative is earthed, The energy stored in the system is (A) \(0.1224 \mu \mathrm{J}\) (B) \(0.2224 \mu \mathrm{J}\) (C) \(0.4224 \mu \mathrm{J}\) (D) \(0.3224 \mu \mathrm{J}\)

Let $\mathrm{P}(\mathrm{r})\left[\mathrm{Q} /\left(\pi \mathrm{R}^{4}\right)\right] \mathrm{r}$ be the charge density distribution for a solid sphere of radius \(\mathrm{R}\) and total charge \(\mathrm{Q}\). For a point ' \(\mathrm{P}\) ' inside the sphere at distance \(\mathrm{r}_{1}\) from the centre of the sphere the magnitude of electric field is (A) $\left[\mathrm{Q} /\left(4 \pi \epsilon_{0} \mathrm{r}_{1}^{2}\right)\right]$ (B) $\left[\left(\mathrm{Qr}_{1}^{2}\right) /\left(4 \pi \in{ }_{0} \mathrm{R}^{4}\right)\right]$ (C) $\left[\left(\mathrm{Qr}_{1}^{2}\right) /\left(3 \pi \epsilon_{0} \mathrm{R}^{4}\right)\right]$
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