Problems and solutions on electric fields are presented for high school and college students. More **practice problems in electric fields** are also provided.

**Problem (1):** The electric field due to charges $q_1=2\,\rm {\mu C}$ and $q_2=32\,\rm {\mu C}$ at distance $16\,\rm {cm}$ from charge $q_2$ is zero. What is the distance between the two charges?

**Solution**:

Since the two charges $q_1$ and $q_2$ are positive, somewhere between them the net electric force must be zero that is at the given point the magnitude of the fields are equal (remember that the electric field of a positive charge at field point is outward). Therefore, we get \[\vec E_{net}=0 \quad \Longrightarrow \quad |\vec E_1|=|\vec E_2|\]

Now use the definition of electric field to evaluate the relation above:

\[k\frac{|q_1|}{r_1^2}=k\frac{|q_2|}{r^{2}} \quad \Longrightarrow \quad \frac{2}{x^{2}}=\frac{32}{16^{2}}\]

Taking square root of the both sides, we obtain \[\frac{1}{x}=\frac{4}{16} \quad \Longrightarrow \quad x=4 \rm {cm}\]

As shown in the figure, the distance of the two charges is $d=x+16=4+16=20\, \rm {cm}$.

**Problem (2):**

In the figure, three equal charges $q_1=q_2=q_3=+4\, \rm {\mu C}$ are located on the perimeter of a sphere of diameter $12\, \rm {cm}$. Find the net electric field, in terms of unit vectors $\hat i,\hat j$ at the center of the sphere.

**Solution**:

**Reasoning**: The electric fields of charges $q_1$ and $q_3$ at the center of the sphere are equal in magnitude and opposite in direction. Since the magnitude of charges are the same $q_1=q_3$ and are located at equal distance from the center so using the definition of the electric field we have

\[E_1=k\frac{|q_1|}{r^2}=E_3 \quad , \quad \vec E_1=-\vec E_3\]

Therefore, the resultant of electric field vectors at point $\rm O$ is, using the superposition principle of fields, equal to the field $\vec E_2$.

\[\vec E_{net,O}=\underbrace{\vec E_1+\vec E_3}_{0}+\vec E_2=\vec E_2\]

But the electric field $\vec E_2$ lies in the forth quadrant along the radius of the sphere which makes the angle of $53^\circ$ relative to the $+x$ axis as shown in the figure. As mentioned already, in such cases we must decompose the vector into its components in $x$ and $y$ directions.

First use the definition of electric field of a point charge to find the magnitude of $\vec E_2$ as

\begin{align*}

\vec E_2&=k\frac{|q_2|}{r^{2}}\\

&=\left(9\times 10^{9}\, \rm {N.\frac{m^2}{C^2}}\right)\frac{4\times 10^{-6}\, \rm C}{\left(\frac{12}{2}\times 10^{-2}\,\rm{cm}\right)^2}\\

&=10^{7}\, \rm{\frac{N}{C}}

\end{align*}

From elementary geometry we can decompose the vector $\vec E_2$ as follows

\begin{align*}

\vec E_2&=\underbrace{|\vec E_2|\,\cos 53^\circ}_{E_{2x}}\left(-\hat i\right)+\underbrace{|\vec E_2|\,\sin 53^\circ}_{E_{2y}}\left(\hat j\right)\\

&=10^{7}(0.6)\left(\hat i\right)+10^{7}(0.8)\left(-\hat j\right)

\end{align*}

By factoring and rearranging the above relation, we get

\begin{align*}

\vec E_2&=10^{7}\left(-0.6 \hat i+0.8 \hat j\right)\\

&=6 \times 10^{6} \hat i-8\times 10^{6} \hat j

\end{align*}

**Problem (3):**

As shown in the figure, the two charges $q_1$ and $q_2$ are fixed at the corners of lower side of a isosceles triangle. If the electric field vector at point $A$ (in SI) is $\vec E_A=\left(7.2 \times 10^{4}\right)\hat i$, determine the type and magnitude of electric charges $q_1$ and $q_2$.

**Solution:**

The solution is straightforward. Use the superposition principle and find two relations between the magnitude of charges. Therefore, we have

\begin{align*}

\vec E_A&=\vec E_1+\vec E_2 \\

\vec E_A&=k \frac{|q_1|}{r_1^2}\hat r_1+k \frac{|q_2|}{r_2^2}\hat r_2

\end{align*}

Where in the second equality we have used the formula of the electric field of a point particle. $r$s are the unit vectors in an arbitrary direction (since we have no knowledge about being positive or negative of charges) that must be found as we proceed. By decomposing unit vectors in $x$ and $y$ directions, and noting that in isosceles triangle $r_1=r_2=d$, we have

\begin{align*}

\vec E_A&=\frac{k}{d^{2}}{|q_1|\left(\cos \alpha\, \hat i+\sin \alpha\, \hat j\right)+|q_2|\left(\cos \alpha \left(-\hat i\right)+\sin \alpha\, \hat j\right)}\\

&=\frac{k}{d^2}{\cos \alpha\, \left(|q_1|-|q_2|\right)\hat i+\sin \alpha\, \left(|q_1|+|q_2|\right)\hat j}

\end{align*}

To decompose the unit vectors we have assumed the charges are positive. Because of electric field at point $A$ is in the positive $x$ direction, so the $j$ component of the right hand side of above must be vanishes and its $i$ components must be equal to the left part as

\begin{gather*}

\vec E_A=\frac{k}{d^2}\,{\cos \alpha \left(|q_1|-|q_2|\right)\hat i+\underbrace{\sin \alpha \left(|q_1|+|q_2|\right)}_{0}\hat j}\\

|q_1|+|q_2|=0\\

7.2\times 10^{4}=\frac{k}{d^2}\, \cos \alpha\, \left(|q_1|-|q_2|\right)

\end{gather*}

The first relation says that the magnitude of charges is opposite each other i.e. $|q_1|=-|q_2|$. From the second equation one can find the \begin{gather*} 2|q_1|\frac{k}{d^2}\,\cos \alpha =7.2\times 10^{4}\\ \Longrightarrow |q_1|=3.6\times 10^{4} \frac{d^2}{k\,\cos \alpha} \end{gather*}

Use the Pythagorean theorem in the left triangle to find the value of $\cos \alpha$.

\begin{gather*} 10^2=6^2+x^2 \quad \Longrightarrow \quad x=8\, \rm {cm}\\ \cos \alpha =\frac{8}{10} \end{gather*}

By substituting the known data, we obtain

\begin{align*}|q_1|&=3.6\times 10^4 \frac{\left(10\times 10^{-2} \rm {m}\right)^{2}}{\left(9\times 10^{9}\right)\left(0.8\right)}\\ &=0.5\times 10^{-9}\, \rm {C}\\ &=0.5\, \rm {nC} \end{align*}

Thus, $|q_1|=-|q_2|=0.5\, \rm {nC}$. Our initial assumption that the charges are positive does not correct. Therefore, we must choose correctly one of them to be positive and the other negative. By choosing $q_1$ to be positive and $q_2$ negative, one can arrive at the right net electric field at point $A$.

**Last Update: 8/7/2020**

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