Question

Two point charges $P = +25\mu\text{C}$ and $Q = -16\mu\text{C}$ are placed $5\text{ cm}$ apart. Find the position of the point at which the resultant electric field is zero:

• A. $25\text{ cm}$ from $Q$ and $20\text{ cm}$ from $P$ on the dipole axis
• B. $20\text{ cm}$ from $Q$ and $25\text{ cm}$ from $P$ on the dipole axis
• C. $2.5\text{ cm}$ from $Q$ and $2.5\text{ cm}$ from $P$ on the dipole axis
• D. $1\text{ cm}$ from $Q$ and $4\text{ cm}$ from $P$ on the dipole axis

Hint:

You can find the null point quickly using the formula: $x = \frac{d}{\sqrt{\frac{q_{\text{large}}}{q_{\text{small}}}} - 1}$. Here, $x = \frac{5}{\sqrt{\frac{25}{16}} - 1} = \frac{5}{\frac{5}{4} - 1} = \frac{5}{\frac{1}{4}} = 20\text{ cm}$ from the smaller charge!

Answer 1

The Correct Option is B

Concept: The electric field produced by a point charge at a distance $r$ is given by $E = \frac{kq}{r^2}$. For two opposite charges, the fields point in opposite directions only at positions outside the region between the two charges along the line connecting them. To completely cancel out, the null point must lie closer to the charge with the smaller magnitude.

Step 1: Identify the correct location zone for the null point.
We have charges $P = +25\mu\text{C}$ and $Q = -16\mu\text{C}$ separated by $d = 5\text{ cm}$. Since $|Q| < |P|$, the point where the fields balance out must lie on the outer side of charge $Q$. Let this point be at a distance $x$ from charge $Q$. Consequently, its total distance from charge $P$ will be $(5 + x)$.

Step 2: Equate field magnitudes to solve for $x$.
\[ E_P = E_Q \implies \frac{k \cdot |q_P|}{(5+x)^2} = \frac{k \cdot |q_Q|}{x^2} \] Substitute the values of the charges: \[ \frac{25}{(5+x)^2} = \frac{16}{x^2} \] Taking the square root on both sides: \[ \frac{5}{5+x} = \frac{4}{x} \] Cross-multiplying to solve: \[ 5x = 4(5 + x) \implies 5x = 20 + 4x \implies x = 20\text{ cm} \]

Step 3: Verify distances from both reference targets.

• Distance from charge $Q = x = 20\text{ cm}$
• Distance from charge $P = 5 + x = 5 + 20 = 25\text{ cm}$