Question

Two balls of charges $q_1$ and $q_2$ initially have exactly same velocity. Both the balls are subjected to same uniform electric field for same time. As a result, the velocity of the first ball is reduced to half of its initial value and its direction changes by $60^\circ$. The direction of the velocity of second ball is found to change by $90^\circ$.

The electric field and initial velocity of the charged particle are inclined at angle

• A. $60^\circ$
• B. $30^\circ$
• C. $90^\circ$
• D. $150^\circ$

Answer 1

Answer: (D)

150^\circ

Answer 2

Let $\vec{v}_0$ be the initial velocity and $\vec{E}$ be the electric field. Let $\theta$ be the angle between $\vec{v}_0$ and $\vec{E}$.

After time $t$, the velocities are $\vec{v}_1 = \vec{v}_0 + \vec{A}_1$ and $\vec{v}_2 = \vec{v}_0 + \vec{A}_2$, where $\vec{A}_1 = \frac{q_1 t}{m_1} \vec{E}$ and $\vec{A}_2 = \frac{q_2 t}{m_2} \vec{E}$. $\vec{A}_1$ and $\vec{A}_2$ are parallel to $\vec{E}$.

For the first ball, $|\vec{v}_1| = v_0/2$ and the angle between $\vec{v}_0$ and $\vec{v}_1$ is $60^\circ$. From the vector triangle $\vec{v}_0 + \vec{A}_1 = \vec{v}_1$, using the law of cosines or dot product, the angle $\alpha_1$ between $\vec{v}_0$ and $\vec{A}_1$ is found to be $150^\circ$ (required for speed reduction).

For the second ball, the angle between $\vec{v}_0$ and $\vec{v}_2$ is $90^\circ$. From $\vec{v}_0 + \vec{A}_2 = \vec{v}_2$, taking the dot product with $\vec{v}_0$, we get $\vec{v}_0 \cdot \vec{v}_0 + \vec{v}_0 \cdot \vec{A}_2 = \vec{v}_0 \cdot \vec{v}_2 = 0$. $v_0^2 + v_0 A_2' \cos \theta = 0$, so $A_2' \cos \theta = -v_0$.

Using the geometric approach with angles of vectors relative to $\vec{v}_0$: Let $\vec{v}_0$ be along $0^\circ$. $\vec{v}_1$ is at $\pm 60^\circ$. $\vec{v}_2$ is at $\pm 90^\circ$. $\vec{A}_1 = \vec{v}_1 - \vec{v}_0$. If $\vec{v}_1$ is at $60^\circ$, direction of $\vec{A}_1$ is $150^\circ$. $\vec{A}_2 = \vec{v}_2 - \vec{v}_0$. If $\vec{v}_2$ is at $90^\circ$, direction of $\vec{A}_2$ is $150^\circ$.

Since $\vec{A}_1 \parallel \vec{A}_2 \parallel \vec{E}$, the direction of $\vec{E}$ must be in the set $\{150^\circ, 210^\circ\}$. The angle $\theta$ between $\vec{v}_0$ (at $0^\circ$) and $\vec{E}$ is the angle of $\vec{E}$ (in $[0^\circ, 180^\circ]$). Thus, $\theta = 150^\circ$.