Q.Two point charges $q_1$ and $q_2$, of magnitude $+10^{-8}\,\text{C}$ and $-10^{-8}\,\text{C}$, respectively, are placed $0.1\,\text{m}$ apart. Calculate the electric fields at points A, B and C shown in Fig. 1.11.

Use $E=\dfrac{k|q|}{r^2}$ with $k=9\times10^{9}$ and $|q|=10^{-8}\,\text{C}$ (so $k|q|=90\,\text{N·m}^2/\text{C}$); add the fields as vectors. Let $q_1(+)$ be on the left and $q_2(-)$ on the right, $0.1\,\text{m}$ apart.

Point A (midpoint, $0.05\,\text{m}$ from each).

$E_1=E_2=\dfrac{90}{(0.05)^2}=3.6\times10^{4}\,\text{N/C}$. $E_1$ points away from $q_1$ (rightward) and $E_2$ toward $q_2$ (rightward), so they add:

Point B ($0.05\,\text{m}$ left of $q_1$; $0.15\,\text{m}$ from $q_2$).

$E_1=3.6\times10^{4}\,\text{N/C}$ (away from $q_1$, leftward); $E_2=\dfrac{90}{(0.15)^2}=4.0\times10^{3}\,\text{N/C}$ (toward $q_2$, rightward). They oppose:

Point C (apex of the equilateral triangle, $0.1\,\text{m}$ from each). …

Each charge produces $E=\dfrac{k|q|}{r^2}$; the net field is the vector sum. A positive charge's field points away from it, a negative charge's field points toward it. Take $q_1=+10^{-8}\,\text{C}$ on the left, $q_2=-10^{-8}\,\text{C}$ on the right, separated by $0.1\,\text{m}$, with $k=9\times10^{9}\,\text{N·m}^2/\text{C}^2$ so that $k|q|=90\,\text{N·m}^2/\text{C}$.

Point A — midpoint. $r=0.05\,\text{m}$ from each:

$E_1$ points away from $q_1$ (to the right); $E_2$ points toward $q_2$ (also to the right). Same direction, so they add:

directed from $q_1$ toward $q_2$.

Point B — $0.05\,\text{m}$ to the left of $q_1$. Then $r_1=0.05\,\text{m}$ and $r_2=0.15\,\text{m}$:

These are opposite, so they subtract:

directed leftward, i.e. away from $q_1$.

Method: Superposition of Electric Fields (Vector Addition)

This problem uses the Principle of Superposition — the net electric field at any point is the vector sum of the fields due to each individual charge.

Step-by-step approach

Step 1: Identify the field contributions

• Each point charge $q$ produces an electric field at a distance $r$ given by:

• Direction: away from positive charge, toward negative charge.

Step 2: For each point (A, B, C):

1. Draw the individual field vectors from $q_1$ and $q_2$ at that point.
2. Calculate the magnitude of each field using the formula above.
3. Resolve vectors into components (if needed).
4. Add components vectorially to get the net field.

Step 3: Apply to each point

Let $k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$

• At point A (midpoint between charges):
  • Both fields point right (away from $+q_1$, toward $-q_2$)
  • $r = 0.05\,\text{m}$ for both
  • $E_1 = E_2 = k \frac{10^{-8}}{(0.05)^2} = 3.6 \times 10^4 \, \text{N/C}$
  • Net field = $E_1 + E_2 = \mathbf{7.2 \times 10^4 \, \text{N/C}}$ (right) …

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting that Electric Field is a Vector

Students often compute the magnitude of the field from each charge and then simply add the numbers — ignoring direction.

Example of the error:

At point B, a student might compute:

$E_1 = \frac{k q_1}{r^2}$ and $E_2 = \frac{k q_2}{r^2}$, then write $E_B = E_1 + E_2$ (wrong!).

Why it's wrong:

Electric field is a vector. Fields from $q_1$ and $q_2$ point in different directions at most points. You must add them using vector addition (head-to-tail or component method).

How to avoid:

• Always draw arrows showing the direction of the field from each charge at the point of interest.
• Remember:
  • Field from a positive charge points away from it.
  • Field from a negative charge points toward it.
• Use a coordinate system and add components: $E_x = E_{1x} + E_{2x}$, $E_y = E_{1y} + E_{2y}$.

Mistake 2: Using the Wrong Distance

Students sometimes use the distance between the two charges ($0.1\,\text{m}$) instead of the distance from each charge to the point.

Example of the error:

For point C (midpoint), a student might write $r = 0.05\,\text{m}$ for both charges — that part is correct. But for point A or B, they might mistakenly use $0.1\,\text{m}$ for both.

Why it's wrong:

Each charge contributes a field based on its own distance to the point. These distances are generally different.

How to avoid:

• Label all distances clearly on the diagram before calculating.
• For each point, write:
  • $r_1 = \text{distance from } q_1 \text{ to the point}$
  • $r_2 = \text{distance from } q_2 \text{ to the point}$
• Use geometry (Pythagoras theorem) if the point is not on the line joining the charges.

Mistake 3: Ignoring the Sign of the Charge in the Direction

Students correctly compute the magnitude $E = \frac{k|q|}{r^2}$, but then assign direction based on the sign incorrectly.

Example of the error:

At point A (to the left of $q_1$), a student might say:

"$q_1$ is positive, so its field points away — that means to the right."

Then for $q_2$ (negative): "Field points toward $q_2$ — that also means to the right."

They then add the magnitudes. This is correct for point A — but the same reasoning applied blindly at point B leads to error.

Why it's wrong:

The direction depends on where the point is relative to the charge, not just the sign of the charge.

How to avoid:

• For each charge, ask: "If I place a positive test charge at this point, which way will it be pushed by this charge?"
• Positive charge → repels the test charge (away from itself).
• Negative charge → attracts the test charge (toward itself).
• Draw the force arrow on the test charge — that's the direction of $\vec{E}$.

Mistake 4: Forgetting to Use $k = 9 \times 10^9$ Correctly

Students sometimes use $k = 9 \times 10^9$ but forget to square the distance or misplace decimal points.

Example of the error:

$E = \frac{9 \times 10^9 \times 10^{-8}}{0.1}$ instead of $\frac{9 \times 10^9 \times 10^{-8}}{(0.1)^2}$.

Why it's wrong:

Coulomb's law for field is $E = \frac{kq}{r^2}$, not $\frac{kq}{r}$.

How to avoid:

• Write the formula explicitly before plugging numbers:

• Check units: $r$ is in metres, so $r^2$ is in m². …