
Two charges $2C$ and $6C$ are separated by a finite distance. If a charge of $-4C$ is added to each of them, then the initial force of \[12\times {{10}^{3}}\]N will change to?
A. \[4\times {{10}^{3}}\] N repulsion
B. \[4\times {{10}^{2}}\]N repulsion
C. \[6\times {{10}^{3}}\]N attraction
D. \[4\times {{10}^{3}}\]N attraction
Answer
573.9k+ views
Hint: A stationary charge produces an electric field around itself. The extent up to which the influence of the field can be felt depends up on the magnitude of the charge. When another charge is there, then the two charges attract or repel each other with a force which is directly proportional to the product of magnitude of two charges and inversely proportional to the square of distance between them. The force can be found out using Coulomb’s law.
Complete step by step answer:
First charge, $A= 2 C$, Second charge, $B= 6C$
Initially force between them is \[12\times {{10}^{3}}\]N, now a charge of $-4 C$ is added to both so, the magnitude of charges are:
$A’= 2-4= -2 C\\
B’= 6-4= 2 C$
So, the new charges are 2 and -2 Coulombs.
Applying Coulomb’s law for first condition:
\[F=k\dfrac{AB}{{{x}^{2}}}\],
where A and B are the magnitude of the two charges and x is the distance between them
$12\times {{10}^{3}}=9\times {{10}^{9}}\times \dfrac{2\times 6}{{{x}^{2}}} \\
\Rightarrow {{x}^{2}}=9\times {{10}^{6}} \\
\Rightarrow x=\sqrt{9\times {{10}^{6}}} \\
\Rightarrow x=3\times 1000 \\
\therefore x=3000m \\$
Now for second case:
$F=k\dfrac{A'B'}{{{x}^{2}}} \\
\therefore F=9\times {{10}^{9}}\times \dfrac{(-2)\times 2}{{{(3000)}^{2}}} \\$
Here minus sign signifies that the force is attractive in nature.
So, the new force is \[4\times {{10}^{3}}\]N and attractive.Hence, the correct option is D.
Note: Electric charge is the basic building block of the living and nonliving substances. The smallest electric charge that could exist is the charge on the electron. Both electron and proton have the same magnitude of charge but of opposite polarity. The charge on protons is positive and the charge on electrons is negative. Both the proton and electron experience electric force. Electric charge is additive in nature and can be added using ordinary law of algebra.
Complete step by step answer:
First charge, $A= 2 C$, Second charge, $B= 6C$
Initially force between them is \[12\times {{10}^{3}}\]N, now a charge of $-4 C$ is added to both so, the magnitude of charges are:
$A’= 2-4= -2 C\\
B’= 6-4= 2 C$
So, the new charges are 2 and -2 Coulombs.
Applying Coulomb’s law for first condition:
\[F=k\dfrac{AB}{{{x}^{2}}}\],
where A and B are the magnitude of the two charges and x is the distance between them
$12\times {{10}^{3}}=9\times {{10}^{9}}\times \dfrac{2\times 6}{{{x}^{2}}} \\
\Rightarrow {{x}^{2}}=9\times {{10}^{6}} \\
\Rightarrow x=\sqrt{9\times {{10}^{6}}} \\
\Rightarrow x=3\times 1000 \\
\therefore x=3000m \\$
Now for second case:
$F=k\dfrac{A'B'}{{{x}^{2}}} \\
\therefore F=9\times {{10}^{9}}\times \dfrac{(-2)\times 2}{{{(3000)}^{2}}} \\$
Here minus sign signifies that the force is attractive in nature.
So, the new force is \[4\times {{10}^{3}}\]N and attractive.Hence, the correct option is D.
Note: Electric charge is the basic building block of the living and nonliving substances. The smallest electric charge that could exist is the charge on the electron. Both electron and proton have the same magnitude of charge but of opposite polarity. The charge on protons is positive and the charge on electrons is negative. Both the proton and electron experience electric force. Electric charge is additive in nature and can be added using ordinary law of algebra.
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