If a body contains $n_1$ electrons and $n_2$ protons, the total amount of charge on the body is
Answer
547.5k+ views
Hint: Both protons and electrons contain charge which but opposite magnitude
Given:
let us see what is given to using the question
The number of electrons on the body are ${n_1}$ and
The number of protons on the body are ${n_2}$
Formula Used:
Here we add the number of electrons and number of protons i.e. \[\left( {{n_1} + {n_2}} \right)e\] to count the number of charges. Where $e$ is the charge
Complete step-by-step solution:
Firstly ,let us understand the concept. There are two kinds of electric charge, positive and negative. On the atomic level, protons are positively charged and electrons are negatively charged.Although the mass of a proton is much larger than that of an electron, the magnitudes of their charges are equal.If an object has more protons than electrons, then the net charge on the object is positive. If there are more electrons than protons, then the net charge on the object is negative. If there are equal numbers of protons and electrons, then the object is electrically neutral.
So let us come to the calculation part, \[\left( {{n_1} + {n_2}} \right)\]
add the number of electrons and protons
Here we must calculate the total amount of charge on the body and consider the path of the charge whether it is negative or positive
As protons carry positive charge”e⁺” and electrons carry negative charge “e⁻”
Now that we know electrons and protons are opposite charges, so the total charge on the body is the algebraic sum of the positive and negative charges on the body,
So $n_1$ will become “$ - {n_1}$ and ${n_2}$ will be “$ + {n_2}$ that is \[\left( {{n_2} - {n_1}} \right)e\]
So the total amount of charge on the body will be \[\left( {{n_2} - {n_1}} \right)e\]
Note: The total charge on the body If there are ${n_1}$ electrons and ${n_2}$ protons will be \[\left| {{n_1} - {n_2}} \right|\]. Because whichever of the two will be lower in number, will be neutralized by the other and the left electrons/ protons will be the reason for the charge on the body.
Given:
let us see what is given to using the question
The number of electrons on the body are ${n_1}$ and
The number of protons on the body are ${n_2}$
Formula Used:
Here we add the number of electrons and number of protons i.e. \[\left( {{n_1} + {n_2}} \right)e\] to count the number of charges. Where $e$ is the charge
Complete step-by-step solution:
Firstly ,let us understand the concept. There are two kinds of electric charge, positive and negative. On the atomic level, protons are positively charged and electrons are negatively charged.Although the mass of a proton is much larger than that of an electron, the magnitudes of their charges are equal.If an object has more protons than electrons, then the net charge on the object is positive. If there are more electrons than protons, then the net charge on the object is negative. If there are equal numbers of protons and electrons, then the object is electrically neutral.
So let us come to the calculation part, \[\left( {{n_1} + {n_2}} \right)\]
add the number of electrons and protons
Here we must calculate the total amount of charge on the body and consider the path of the charge whether it is negative or positive
As protons carry positive charge”e⁺” and electrons carry negative charge “e⁻”
Now that we know electrons and protons are opposite charges, so the total charge on the body is the algebraic sum of the positive and negative charges on the body,
So $n_1$ will become “$ - {n_1}$ and ${n_2}$ will be “$ + {n_2}$ that is \[\left( {{n_2} - {n_1}} \right)e\]
So the total amount of charge on the body will be \[\left( {{n_2} - {n_1}} \right)e\]
Note: The total charge on the body If there are ${n_1}$ electrons and ${n_2}$ protons will be \[\left| {{n_1} - {n_2}} \right|\]. Because whichever of the two will be lower in number, will be neutralized by the other and the left electrons/ protons will be the reason for the charge on the body.
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