Question

Charges $Q_1$ and $Q_2$ are at points $A$ and $B$ of a right angle triangle $OAB$ (see figure). The resultant electric field at point $O$ is perpendicular to the hypotenuse, then $Q_1/Q_2$ is proportional to

• A. a/b
• B. b/a
• C. a^2/b^2
• D. b^2/a^2

Answer 1

Answer: (A)

a/b

Answer 2

Let $O$ be the origin $(0,0)$, $A$ be at $(a,0)$ and $B$ be at $(0,b)$. Charge $Q_1$ is at $A$ and $Q_2$ is at $B$. The electric field at $O$ due to $Q_1$ is $\vec{E}_1 = \frac{k Q_1}{a^2}(-\hat{i})$. The electric field at $O$ due to $Q_2$ is $\vec{E}_2 = \frac{k Q_2}{b^2}(-\hat{j})$. The resultant electric field at $O$ is $\vec{E} = \vec{E}_1 + \vec{E}_2 = -\frac{k Q_1}{a^2}\hat{i} - \frac{k Q_2}{b^2}\hat{j}$.

The hypotenuse $AB$ connects points $A(a,0)$ and $B(0,b)$. The vector representing the hypotenuse is $\vec{AB} = -a\hat{i} + b\hat{j}$.

Since $\vec{E}$ is perpendicular to $\vec{AB}$, their dot product is zero: $\vec{E} \cdot \vec{AB} = 0$ $\left(-\frac{k Q_1}{a^2}\hat{i} - \frac{k Q_2}{b^2}\hat{j}\right) \cdot (-a\hat{i} + b\hat{j}) = 0$ $\frac{k Q_1}{a} - \frac{k Q_2}{b} = 0$ $\frac{Q_1}{a} = \frac{Q_2}{b}$ Therefore, $\frac{Q_1}{Q_2} = \frac{a}{b}$.