Question

If the electrical force between two charges is $200N$and we increase $10\% $ of the charge on one of the charges and decrease $10\% $ on the other, then the electrical force between them for the same distance becomes
A. $200N$
B. $202N$
C. $198N$
D. $199N$

Answer 1

Hint: To find the force between two charges, Coulomb’s law is used. The force between two charges is inversely proportional to the square of the distance between them, this is known as Coulomb’s law. A constant is introduced to remove the proportionality sign. Using this relation, the given problem can be solved.

Formula used:
$ \Rightarrow F = \dfrac{{K{q_1}{q_2}}}{{{r^2}}}$
Where,
$F$ is the force, ${q_1},{q_2}$ are two charges, $r$ radius, and $K$ is the constant.

Complete step by step solution:
Given two charges has the electric force of $200N$. To find the electric forces between them when we increase or decrease the charges by $10\% $. Consider the given diagram.

To solve the given problem, consider Coulomb's law. The force between two charges is inversely proportional to the square of the distance between them, this is known as Coulomb’s law. A constant is introduced to remove the proportionality sign. The mathematical representation of the law:
$ \Rightarrow F = \dfrac{{K{q_1}{q_2}}}{{{r^2}}}$
Where,
$F$ is the force, ${q_1},{q_2}$ are two charges, $r$ radius, and $K$ is the constant.
There are two charges given. Find the charges on the two points separately. Consider the $10\% $ as a decimal number as $0.1$
$ \Rightarrow {q_1}^\prime = {q_1} + \left( {0.1} \right){q_1}$
$ \Rightarrow {q_1}^\prime = 1.1{q_1}$
Similarly for other charge,
$ \Rightarrow {q_2}^\prime = {q_2} + \left( {0.1} \right){q_2}$
$ \Rightarrow {q_2}^\prime = 0.9{q_2}$
Calculate the forces on the charge. The formula is,
$ \Rightarrow F = \dfrac{{K{q_1}{q_2}}}{{{r^2}}} = 200N$
And,
$ \Rightarrow F' = \dfrac{{K\left( {{{q'}_1}} \right){q_1}\left( {{{q'}_2}} \right){q_2}}}{{{r^2}}}$
Substitute the values of the point charges.
$ \Rightarrow F' = \dfrac{{K\left( {1.1} \right){q_1}\left( {0.9} \right){q_2}}}{{{r^2}}}$
Simplify the equation,
$ \Rightarrow F' = \left( {1.1} \right) \times \left( {0.9} \right)\dfrac{{K{q_1}{q_2}}}{{{r^2}}}$
The value of $\dfrac{{K{q_1}{q_2}}}{{{r^2}}}$ is $200N$
$ \Rightarrow F' = \left( {1.1} \right) \times \left( {0.9} \right) \times 200N$
Simplify using multiplication.
$ \Rightarrow F' = 0.99 \times 200N$
$ \Rightarrow F' = 198N$
Therefore, the electrical force between them for the same distance becomes $198N$.
Hence, the option $\left( C \right)$ is the correct option.

Note:
There are two types of electrical charges. One is positive and another is negative. Between the two-point charges, the coulomb’s force between them is attractive when both the charges have opposite signs. If the force is negative, the coulomb’s force between them is repulsive when both have the same sign.