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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

Charges \(+q\) and \(-q\) are placed at point \(A\) and \(B\) respectively which are a distance \(2 \mathrm{~L}\) apart, \(\mathrm{C}\) is the midpoint between \(\mathrm{A}\) and \(\mathrm{B}\). The work done in moving a charge \(+Q\) along the semicircle \(C R D\) is \(\ldots \ldots\) (A) $\left[(\mathrm{qQ}) /\left(2 \pi \mathrm{\epsilon}_{0} \mathrm{~L}\right)\right]$ (B) \(\left[(-q Q) /\left(6 \pi \in_{0} L\right)\right]\) (C) $\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]$ (D) \(\left[(q Q) /\left(4 \pi \in_{0} L\right)\right]\)

Two metal plate form a parallel plate capacitor. The distance between the plates is \(\mathrm{d}\). A metal sheet of thickness \(\mathrm{d} / 2\) and of the same area is introduced between the plates. What is the ratio of the capacitance in the two cases? (A) \(4: 1\) (B) \(3: 1\) (C) \(2: 1\) (D) \(5: 1\)

A small conducting sphere of radius \(r\) is lying concentrically inside a bigger hollow conducting sphere of radius \(R\). The bigger and smaller sphere are charged with \(\mathrm{Q}\) and \(\mathrm{q}(\mathrm{Q}>\mathrm{q})\) and are insulated from each other. The potential difference between the spheres will be \(\ldots \ldots\) (A) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(q / r)-(Q / R)]\) (B) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(q / r)-(q / R)]\) (C) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(Q / R)+(q / r)]\) (D) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(\mathrm{q} / \mathrm{R})-(\mathrm{Q} / \mathrm{r})]$

Three charges \(2 q,-q,-q\) are located at the vertices of an equilateral triangle. At the centre of the triangle. (A) The Field is Zero but Potential is non - zero (B) The Field is non - Zero but Potential is zero (C) Both field and Potential are Zero (D) Both field and Potential are non - Zero

The electric Potential at a point $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})\( is given by \)\mathrm{V}=-\mathrm{x}^{2} \mathrm{y}-\mathrm{x} \mathrm{z}^{3}+4\(. The electric field \)\mathrm{E}^{\boldsymbol{T}}$ at that point is \(\ldots \ldots\) (A) $i \wedge\left(2 \mathrm{xy}+\mathrm{z}^{3}\right)+\mathrm{j} \wedge \mathrm{x}^{2}+\mathrm{k} \wedge 3 \mathrm{xz}^{2}$ (B) $i \wedge 2 \mathrm{xy}+\mathrm{j} \wedge\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)+\mathrm{k} \wedge\left(3 \mathrm{xy}-\mathrm{y}^{2}\right)$ (C) \(i \wedge z^{3}+j \wedge x y z+k \wedge z^{2}\) (D) \(i \wedge\left(2 x y-z^{3}\right)+j \wedge x y^{2}+k \wedge 3 z^{2} x\)
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