Electric Charge: Sample Problems with Solutions

Electric charge problems with full explanations are provided for high school physics. Here, we will learn how to find the number of electrons transferred during a charging process, such as induction and conduction. 


Electric Charge Problems


Problem (1): Once a glass rod is rubbed with a silk cloth, it is observed that a positive charge of $+3\,\rm nC$ is transferred to the glass. In the rubbing process, have protons been added to the rod, or have electrons been removed from it?

Solution: in this experiment, we observe that the glass rod has been positively charged by $+3\,\rm nC$. This means that the glass rod must have an excess of positive charges (or deficiency of electrons), and the silk must have received that amount of electrons exited from the glass.  

Notice that in the electrically charging process, such as conduction and induction, we assume that only electrons can be transferred freely from the objects, not the positive constituents of atoms (protons). 

Thus, the electrons are removed from the glass rod and added to the fur cloth. 


Problem (2): when a rubber rod is rubbed with a wool cloth, it is observed that the rod becomes negatively charged. What happens to the charge of the wool cloth and why? 

Solution: the rod initially is neutral, meaning it has an equal number of electrons and protons. After rubbing with wool, it becomes negatively charged. That tells us that number of electrons in the rod must be more than protons.

These number of excess electrons must be removed from wool, leaving it as a positively charged object.

Problem (3): A small metallic sphere hangs from the ceiling by a plastic (insulating) thread. When a positively charged rod brings near the sphere, it is attracted by the sphere. What is the charge of the ball? 

Solution: Charges of the same signs repel each other while charges of opposite signs attract. This is one of the fundamental observations in the electrostatic. 

In this case, a positively charged rod is attracted by a small metallic sphere, so we conclude that the sphere must have a negative charge.


Problem (4): A nail with a mass of $23\,\rm g$ is charged with $-3.6\times 10^{-23}\,\rm C$. How many excess electrons are there in the nail? (The fundamental charge is $e=1.6\times 10^{-19}\,\rm C$. 

Solution: The electric charge is quantized. The electric charge of an object depends on the number of electrons it gains or loses during a physical process, such as charging with conduction and/or induction. 

In these cases, the number of electrons transferred is calculated by the following formula \[q=\pm ne\] where the plus sign is for when the object loses electrons and the negative for gaining electrons. $e=1.6\times 10^{-19}\,\rm C$ is the basic or fundamental electric charge.

In this question, the nail has a negative charge so it must have excess electrons whose number is calculated as follows \begin{align*} n&=-\frac{q}{e} \\\\ &=-\frac{-3.6\times 10^{-23}}{1.6\times 10^{-19}} \\\\&=\boxed{2.25\times 10^4} \end{align*} Therefore, if $2.25\times 10^4$ electrons are added to this nail by induction or conduction, it will ended up with a net charge of $q=-3.6\times 10^{-23}\,\rm C$. 

Problem (5): How many electrons can make up a charge of $q=+24\,\rm \mu C$? 

Solution: here, we have an object with a charge of $q=+24\,\rm \mu C$. Generally, every object is neutral in nature, meaning that the number of its protons (positive charges) equals the number of its electrons (negative charges). Once They lose some of their electrons, they become positively charged objects. The total number of transferred electrons is related to the basic electric charge $e=1.6\times 10^{-19}\,\rm C$ by the formula $q=\pm ne$. 

Above, the plus is for a positively charged object, and the minus sign is for a negatively charged object.

In this electric charge problem, the object has a positive charge, so the correct formula is $q=+ne$. 

Therefore, an object with a total charge of $q+24\,\rm \mu C$ is made up of the following number of electrons \begin{align*} n&=\frac qe \\\\ &= \frac{24\times 10^{-6}\,\rm C}{1.6\times 10^{-19}\,\rm C} \\\\ &=\boxed{15\times 10^{13}} \end{align*} note that $\rm \mu C=10^{-16}\,C$. 


Problem (6): How many excess electrons are in a tiny droplet of oil of charge of $-6\times 10^{-6}\,\rm C$?

Solution: the oil droplet, originally, must be neutral, but with some method, it has been negatively charged, say by induction. The number of electrons it gains is calculated by the formula $q=-ne$. Therefore, solving for $n$ gives us \begin{align*} n&=-\frac qe\\\\ &=-\frac{-6\times 10^{-6}}{1.6\times 10^{-19}}\\\\&=\boxed{3.75\times 10^{13}} \end{align*} 


Problem (7): Two isolated and identical conducting spheres are charged with charges of $+8\,\rm \mu C$ and $-2\,\rm \mu C$, respectively. We briefly contact these two spheres with each other and separate them again. Determine the charge on each sphere.

Solution: According to the conservation of the electric charge, the net charge of two objects before and after touching must be equal. \[q_{before}=q'_{after}\] The net charge on the spheres is the algebraic sum of the charges each has. Before touching, the net charge of both spheres is \begin{align*} q_{net}&=q_1+q_2 \\&=8+(-12) \\&=-4\,\rm \mu C \end{align*} After touching the two spheres and then separating, they must have the same charges as before \[q'_{after}=-4\,\rm \mu C\] Since the two spheres are identical in size, they must contain an equal amount of charges after separation. Therefore, each sphere has a charge of $-2\,\rm \mu C$, after separation. 


Problem (8): An object has a charge of $-1.6\,\rm \mu C$. Is the number of protons of this object more or less than the number of electrons? How many?

Solution: The charge of the object is negative, so it has gained electrons and consequently, its electrons are more than the protons. To understand the object how many protons have less than electrons, we must use the formula $q=-ne$ and solve for $n$. \begin{align*} n&=-\frac{q}{e}\\\\&=-\frac{-1.6\times 10^{-6}}{1.6\times 10^{-19}}\\\\&=\boxed{10^{13}} \end{align*} This means that the number of protons of the object is $10^{13}$ times less than the electrons.


Problem (9): An object has been electrically charged by rubbing. Which of the following can be the total electric charge of the object? $8\,\rm \mu C$ or $4\,\rm \mu C$. 

Solution: the electric charge is a quantized quantity, meaning that they come in discrete packets that are integral multiples of the fundamental electric charge $e=1.6\times 10^{-19}\,\rm C$. This statement is illustrated in the formula $q=ne$. 

For the object to have a charge of $q=8\,\rm \mu C$, it must gains or loses some electrons as much as the following \[n=\frac{q}{e}=\frac{8\times 10^{-19}}{1.6\times 10^{-19}}=5\] which is acceptable as it is a whole number. 

Now assume the object's charge wants to be $q=4\,\rm \mu C$. Thus, the number of excess or deficit electrons is calculated as \[n=\frac{4\times 10^{-19}}{1.6\times 10^{-19}}=2.5\] This is not acceptable, since there are no $2.5$ times basic electric charges in nature. 

Problem (10): A plastic rod is rubbed with fur until it acquires a net charge of $-6\,\rm nC$. Assuming that only electrons can be transferred between them,
(a) Were electrons removed from the rod or added to it?  
(b) How many electrons were transferred during this charging process? 

Solution: (a) As a result of rubbing a neutral plastic rod with fur, it has been negatively charged. This tells us that it must has been acquire some electrons so that now it has an excess of electrons with negative charges. 

(b) We are given the amount of charge as $q=-6\,\rm nC$. Applying the electric charge formula $q=-ne$ and solving for the number of electrons $n$, yields \begin{align*} n&=-\frac{q}{e} \\\\&=-\frac{-6\times 10^{-9}}{1.6\times 10^{-19}} \\\\&=3.75\times 10^{10} \end{align*} Therefore, around $\boxed{3.75\times 10^{10}}$ electrons are removed from fur and added to the plastic rod. 

Problem (11): Two identical conducting spheres, each having charges of $q_1=-8\,\rm \mu C$ and $q_2=24\,\rm \mu C$, are connected by a conducting wire and then separated. 
(a) What would be the electric charges of each sphere?
(b) Which sphere and how many electrons does it gain?

Solution: (a) The spheres are identical (same size and material), so according to the conservation of electric charge, the net charge before and after connecting must be equal. \[q=q'\] On the other hand, the net charge is also defined as the algebraic sum of charges. Therefore, the net charge after connecting is \begin{align*} q'&=q_1+q_2 \\&=-8+24\\&=16\,\rm \mu C \end{align*} This is the charge that the spheres have just after connecting to each other. Once separated, since they are identical, they each have half of the total charge after connecting, i.e., \[q_{each}=\frac{q'}{2}=8\,\rm C\] 
(b) In the previous part, we saw that each sphere, after connecting and separation, has a charge of $8\,\rm \mu C$. The sphere $2$ initially has a positive charge of $+24\,\rm \mu C$, and will be ended up with the charge $+8\,\rm \mu C$. Thus, it must have received $-16\,\rm \mu C$ of charge to neutralize some of its original charge of $+24\,\rm \mu C$. 

Consequently, $-16\,\rm \mu C$ of charge must be transferred from sphere $1$ to sphere $2$. Now, we use the formula $q=-ne$ to find the number of transferred electrons  \begin{align*} n&=-\frac{q}{e}\\\\&=-\frac{-16\times 10^{-6}}{1.6\times 10^{-19}}\\\\&=10^{14} \end{align*} Therefore, sphere $2$ receives $10^{14}$ electrons from sphere $1$. 

Note that if you are confused with the sign of the equation $q=\pm ne$, always remember that the number of positive/negative transferred charges must be a positive number.


Problem (12): Two small spheres separated by $20\,\rm cm$ have an equal amount of charge. How many excess electrons must be carried by each sphere to produce a net repulsive electric force of $5.66\times 10^{-21}\,\rm N$? 

Solution: according to the solved problems on Coulomb's law, the electric force between two electrically charged objects placed at a distance of $d$ is calculated using the following formula \[F=k\frac{qq'}{d^2}\] where $k=8.99\times 10^9\,\rm N\cdot m^2/C^2$ is the coulomb constant. 

We are given two spheres with equal charges, $q=q'$. Therefore, by substituting the numerical values into the above formula and solving for the unknown charge $q$, we will have \begin{align*} q &=\sqrt{\frac{Fd^2}{k}}\\\\&=\sqrt{\frac{(5.66\times 10^{-21})(0.20)^2}{8.99\times 10^9}}\\\\&=1.586\times 10^{-16}\,\rm C \end{align*} Now that the amount of electric charge on each sphere is found, the number of electrons $n_e$ is calculated as follows \begin{align*} n_e&=\frac{q}{e}\\\\&=\frac{1.586\times 10^{-16}}{1.6\times 10^{-19}} \\\\&=\boxed{990} \end{align*} Therefore, each sphere has $990$ excess electron. 

Problem (13): A rod, having a negative charge of $-6\,\rm nC$, is brought near and touched with a metallic sphere of charge $+4\,\rm nC$. After touch, $8.2\times 10^9$ electrons are transferred from the rod to the sphere. Now, what are the charges of the sphere and the rod? 

Solution: before touching the net charge of both objects is $q=-6+4=-2\,\rm nC$. According to the conservation of electric charge, after touching, the objects must have the same amount of net charges. 

We are given the number of transferred electrons, so first, find the charge corresponding to this number of electrons. \begin{align*} q_{trans}&=ne\\\\&=(8.2\times 10^9)(1.6\times 10^{-19}) \\\\&=1.312\times 10^{-9}\,\rm C \\\\&=1.312\,\rm nC \end{align*} Thus, in this process, $q_{trans}=1.312\,\rm nC$ electric charge is removed from the rod and transferred to the sphere. Thus, the rod must have the following amount of charge \[ q_{rod}=-6+1.312=-4.688\,\rm nC \] The sphere also receives $q_{trans}$ and will ended up with the following charge \[q_{sphere}=+4-1.312=2.688\,\rm nC\]


Problem (14): What is the total charge of $76\,\rm kg$ of electrons? Using this, can you estimate the number of electrons in your body?

Solution: The mass of one electron is $9.11\times 10^{-31}\,\rm kg$, so the number of electrons in $76\,\rm kg$ is calculated as follows \begin{align*} n_e&=\frac{76}{9.11\times 10^{-31}}\\\\ &=8.34\times 10^{31}\, electrons \end{align*} This number of electrons can make an electric charge of \begin{align*} q&=-ne\\\\&=-(8.34\times 10^{31})(1.6\times 10^{-19}) \\\\&=-1.33\times 10^{13}\,\rm C \end{align*} Note that the average weight of a human body is about $76\,\rm kg$ and normally your body has no net charge. Therefore, these computations could be a rough estimate of the number of electrons in your body. 

Problem (15): Assuming a U.S. penny is made entirely of copper and has a mass of $3.11\,\rm g$, 
(a) how many electrons would have to be removed from it to leave a net charge of $1\times 10^{-7}\,\rm C$ on it?
(b) What fraction of all electrons does this number correspond to?
(the atomic mass number of copper is $63.54\,\rm g/mol$). 

Solution: originally all objects in nature are neutral, meaning they have an equal number of electrons, protons, and neutrons. If a certain number of electrons are removed from an object, it will experience a decrease in its negative charge constituents and subsequently an increase in the positive charges. 

(a) The number of electrons that must be removed to get a net charge of $1\times 10^{-7}\,\rm C$ is determined by the following equation \begin{align*} n&=\frac{q}{e}\\\\ &=\frac{1\times 10^{-7}}{1.6\times 10^{-19}}\\\\&=6.2\times 10^{11} \end{align*} 
(b) In the first step, we must find the total number of electrons (or protons) in a $3.11-\rm g$ penny made of copper.

The atomic mass number of copper tells us that one mole of copper (or $6.02\times 10^{23}$ of copper atoms) has a mass of $63.54\,\rm g$.
 Thus, a simple proportional relationship gives us the number of copper atoms in a $3.11-\rm g$ penny. \begin{align*} N_{Cu}&=\frac{3.11\times (6.02\times 10^{23}}{63.54}\\\\&=2.95\times 10^{22} \end{align*} This is the number of copper atoms in a penny. On the other hand, each copper atom is made of $29$ electrons and protons. Thus, the number of electrons in a penny is \begin{align*} n_e&=29\times (2.95\times 10^{22}) \\\\&=8.55\times 10^{23}\end{align*} Hence, $6.2\times 10^{11}$ electrons must be removed from a neutral penny with a total number of electrons of $2.95\times 10^{22}$ to get a electrically charged penny. Thus, the fraction of electrons that must be removed is computed as below \[f=\frac{6.2\times 10^{11}}{8.55\times 10^{23}}=7.3\times 10^{-13}\] which is a very small fraction of the total number of electrons that a penny can have.

Problem (16): Two small aluminum spheres, each having a mass of $0.010\,\rm kg$ are at a distance of $55\,\rm cm$. 
(a) How many electrons does each sphere have?
(b) How many electrons must be removed from one sphere and added to the other to produce an attractive force of $1\times 10^4\,\rm N$ between the spheres?
(The atomic mass of Al is $26.982\,\rm g/mol$, and its atomic number is $13$)

Solution: the atomic mass of Al, $26.982\,\rm g/mol$, indicates that one mole (or $6.02\times 10^{23}$ atoms) of aluminum has $26.982\,\rm g$. Initially, the spheres are neutral, so the number of electrons and protons are the same, i.e., $n_e=n_p$. 

(a) Using a simple proportional relationship, we can find the number of aluminum atoms present in $\rm 0.010\,kg$ as below \begin{align*} N_{Al}&=\frac{0.010\times (6.02\times 10^{23})}{26.982}\\\\&=2.23\times 10^{20} \end{align*} On the other hand, we are given the aluminum atomic number, denoted by $Z$, which is $13$. The atomic number tells us one element, say aluminum, how many electrons and protons have. 

In this case, the atomic number $13$ indicates that each aluminum atom has $13$ electrons and $13$ protons. Therefore, the total number of electrons present in $2.23\times 10^{20}$ Al atoms is simply computed as follows \begin{align*} n_e&=N_{Al}\cdot Z \\\\ &=(2.23\times 10^{20})(13) \\\\&=2.9\times 10^{21} \end{align*} Therefore, there is around $\boxed{2.9\times 10^{21}}$ electrons in $0.010\,\rm kg$ of aluminum. 

(b) Once a certain number of electrons is removed from a neutral object and added to another object, the first object becomes positively charged (due to deficiency of electrons) and the other negatively charged (due to the excess of electrons). 

In this process, both objects have an equal number of charged particles ($q_1=q_2$). In the electric force problems, we learned that when these two charged objects are at a distance of $d$, according to Coulomb's law, they exert an electric force on each other whose magnitude is found as \[F=k\frac{q_1 q_2}{d^2}\] In this question, $q_1=q_2=q$ which is unknown and must be found. Substitute the known values into this equation and solve for $q$ as follows \begin{align*} q&=\sqrt{\frac{Fd^2}{k}} \\\\ &=\sqrt{\frac{(1\times 10^4)(0.55)^2}{8.99\times 10^9}} \\\\ &=5.8\times 10^{-4}\,\rm C \end{align*} This is the total charge must be transferred from one to the other which corresponds to the following number of electrons \begin{align*} n&=\frac{q}{e} \\\\&=\frac{5.8\times 10^{-4}}{1.6\times 10^{-19}} \\\\ &=3.625\times 10^{15} \end{align*} Therefore, if $\boxed{3.625\times 10^{15}}$ electrons are removed from one sphere and added to the other, an attractive force of $10^4\,\rm N$ is produced between the spheres. This amount of force is pretty large compared to the previous problem. 

We can also find the fraction of transferred electrons to the total electrons (part (a)) existed in each sphere as \begin{align*} f&=\frac{3.625\times 10^{15}}{2.9\times 10^{21}}\\\\ &=1.25\times 10^{-6} \end{align*} It is approximately one millionth of the total electrons present in each sphere. A very small number.

Problem (17): A current of $0.25\,\rm A$ passing through a conducting wire for $4.5\,\rm min$. Approximately how many electrons flow through the wire? 

Solution: from electric current problems recall that the electric current in a wire is defined as \[I=\frac{\Delta q}{\Delta t}\] where $\Delta q$ is the amount of charge that passes through a cross-sectional area of the wire in a time interval $\Delta t$. 

We are given the current through the wire $I=2.5\,\rm A$ in a time interval $\Delta t=2.5\,\rm min$. Now, we must find the amount of electric charge that is moved during this time interval. \begin{align*} \Delta q&=I\Delta t\\ &=(0.25)(270) \\&=67.5\,\rm C \end{align*} where we used this fact that a minute is $60\,\rm s$. 

Finally, use the equation $q=ne$ to find the number of electrons transferred through the wire in that time interval. \begin{align*} n&=\frac{q}{e} \\\\ &=\frac{67.5}{1.6\times 10^{-19}} \\\\&=\boxed{4.2\times 10^{20}} \end{align*} 

Problem (18): (a) How many protons are there in a $15.6-\rm g$ gold ring? 
(b) Assuming the ring has no net charge, how many electrons are there in it? The atomic mass and number of gold are $197\,\rm g/mol$, and $79$, respectively. 

Solution: the atomic mass number of gold is $\rm 197\, g/mol$, meaning that $\rm 197\,g$ of gold equals one mole. Recall that one mole of any substance also has $6.02\times 10^{23}$ units of that substance (such as atoms, ions, and molecules) which is known as the Avogadro number.  

(a) Thus, $197\,\rm g$ of gold has $6.02\times 10^{23}$ protons. The number of protons in $15.6\,\rm g$ of gold is also calculated simply by a proportional relationship \begin{align*} n&=\frac{15.6\times(6.02\times 10^{23})}{197} \\\\&=4.76\times 10^{22}\end{align*} 
(b) ''No neat charge'' means that the number of positive charges (protons) is equal to the number of negative charges (electrons). Hence, this ring has $4.76\times 10^{22}$ electrons. 

Practice Problem (19): How many positive charges are there in 1 gram of water? 

Practice problem (20): How many positive charges are there in one mole of water? 


In this complete guide, we solved some problems on electric charge. 

Author: Dr. Ali Nemati
Published: Sep 27, 2022