Question

The distance between two charges \( q_1 = +2 \,\mu\text{C} \) and \( q_2 = +8 \,\mu\text{C} \) is \( 15 \text{ cm} \). Calculate the distance from the charge \( q_1 \) to the points on the line segment joining the two charges where the electric field is zero.

• A. $1 \text{ cm}$
• B. $2 \text{ cm}$
• C. $3 \text{ cm}$
• D. $4 \text{ cm}$
• E. $5 \text{ cm}$

Hint:

For two like charges, the neutral point always lies between them[ 14]. For unlike charges, it lies outside the segment, closer to the charge with the smaller magnitude.

Answer 1

Answer: (E)

Concept: For the net electric field to be zero at a point between two positive charges, the individual fields produced by each charge must be equal in magnitude and opposite in direction.

Step 1: {Set up the distances for the neutral point.}
Let the distance from $q_1$ be $x \text{ cm}$. Then the distance from $q_2$ is $(15 - x) \text{ cm}$.

Step 2: {Apply the electric field formula for equilibrium.}
Using $E = \frac{kQ}{r^2}$, we set $E_1 = E_2$: $$\frac{k q_1}{x^2} = \frac{k q_2}{(15 - x)^2}$$

Step 3: {Substitute the charge values and simplify.}
$$\frac{2}{x^2} = \frac{8}{(15 - x)^2} \implies \frac{1}{x^2} = \frac{4}{(15 - x)^2}$$ Taking the square root of both sides: $$\frac{1}{x} = \frac{2}{15 - x}$$

Step 4: {Solve for $x$.}
$$15 - x = 2x \implies 3x = 15 \implies x = 5 \text{ cm}$$