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Chapter 11: Problem 1525

Two point charges repel each other with a force of \(100 \mathrm{~N}\). One of the charges is increased by \(10 \%\) and other is reduced by \(10 \%\). The new force of repulsion at the same distance would be \(\ldots \ldots \mathrm{N}\). (A) 121 (B) 100 (C) 99 (D) 89

Short Answer

Expert verified
The new force of repulsion at the same distance after altering the charges is \(99 N\), which corresponds to the answer (C).

Step by step solution

01

Write down the formula for Coulomb's law

The formula for the force between two point charged objects at distance r can be given by Coulomb's Law: \[F = k \frac{q_1 q_2}{r^2}\] Where \(F\) is the force, \(k\) is the Coulomb's constant (\(8.99 \times 10^9 N m^2 C^{-2}\)), \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between them.
02

Find the relation between initial charges and force

Let the initial charges be \(q_1\) and \(q_2\), and the force between them is 100 N. According to Coulomb's law, the initial force can be written as: \[F_1 = k \frac{q_1 q_2}{r^2} = 100 N\]
03

Modify the charges and write the new force equation

Now, we increase one charge by 10% and decrease the other charge by 10%. The new charges will be: \[q_1' = 1.1q_1\] \[q_2' = 0.9q_2\] Now, we can write the equation for the new force between the charges at the same distance \(r\): \[F_2 = k \frac{q_1' q_2'}{r^2}\]
04

Calculate the new force

Replacing the new charges in the above equation, we get: \[F_2 = k \frac{1.1q_1 * 0.9q_2}{r^2}\] \[F_2 = k \frac{0.99q_1 q_2}{r^2}\] We know that the initial force equation is \(F_1 = k \frac{q_1 q_2}{r^2} = 100 N\). Thus, we can write the new force equation in terms of the initial force equation: \[F_2 = 0.99F_1\] \[F_2 = 0.99(100 N)\]
05

Find the new force value

Calculating the new force, we get the final value: \[F_2 = 99 N\] So, the new force of repulsion at the same distance after altering charges is 99 N, which corresponds to the answer (C).

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Most popular questions from this chapter

Three identical spheres each having a charge \(\mathrm{q}\) and radius \(R\), are kept in such a way that each touches the other two spheres. The magnitude of the electric force on any sphere due to other two is \(\ldots \ldots \ldots\) (A) $(\mathrm{R} / 2)\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{5} / 4)(\mathrm{q} / \mathrm{R})^{2}$ (B) $\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{2} / 3)(\mathrm{q} / \mathrm{R})^{2}$ (C) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{3} / 4)(\mathrm{q} / \mathrm{R})^{2}$ (D) $-\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{3} / 2)(\mathrm{q} / \mathrm{R})^{2}$

Two positive point charges of \(12 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are \(10 \mathrm{~cm}\) apart each other. The work done in bringing them $4 \mathrm{~cm}$ closer is .... (A) \(5.8 \mathrm{~J}\) (B) \(13 \mathrm{eV}\) (C) \(5.8 \mathrm{eV}\) (D) \(13 \mathrm{~J}\)

The electric Potential \(\mathrm{V}\) at any Point $0(\mathrm{x}, \mathrm{y}, \mathrm{z}\( all in meters \))\( in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}\( volt. The electric field at the point \)(1 \mathrm{~m}, 0.2 \mathrm{~m})\( in volt meter is \)\ldots \ldots .$ (A) 8 , along negative \(\mathrm{x}\) - axis (B) 8 , along positives \(\mathrm{x}\) - axis (C) 16 , along negative \(\mathrm{x}\) -axis (D) 16 , along positives \(\mathrm{x}\) -axis

A thin spherical shell of radius \(R\) has charge \(Q\) spread uniformly over its surface. Which of the following graphs, figure most closely represents the electric field \(\mathrm{E}\) (r) produced by the shell in the range $0 \leq \mathrm{r}<\infty\(, where \)\mathrm{r}$ is the distance from the centre of the shel1.

An electrical technician requires a capacitance of \(2 \mu \mathrm{F}\) in a circuit across a potential difference of \(1 \mathrm{KV}\). A large number of $1 \mu \mathrm{F}$ capacitors are available to him, each of which can withstand a potential difference of not than \(400 \mathrm{~V}\). suggest a possible arrangement that requires a minimum number of capacitors. (A) 2 rows with 2 capacitors (B) 4 rows with 2 capacitors (C) 3 rows with 4 capacitors (D) 6 rows with 3 capacitors
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