Question

Statement I: If three point charges are placed at the corners of an equilateral triangle, then the system can never be in equilibrium. Statement II: the system of three point charges lying on a straight line can be in equilibrium

• A. Statement I is true and Statement II is true
• B. Statement I is true and Statement II is false
• C. Statement I is false and Statement II is true
• D. Statement I is false and Statement II is false

Answer 1

Answer: (D)

Statement I is false and Statement II is false

Answer 2

Analysis of Statement I:

For a system of three point charges to be in equilibrium, the net force on each charge must be zero. Let the three charges be $q_1, q_2, q_3$ placed at the vertices of an equilateral triangle of side length $a$.

Consider the charge $q_1$. It experiences forces from $q_2$ and $q_3$. For $q_1$ to be in equilibrium, the vector sum of the forces from $q_2$ and $q_3$ must be zero: $\vec{F}_{12} + \vec{F}_{13} = 0$. If the charges are $q_A, q_B, q_C$ at vertices A, B, C, then for $q_A$ to be in equilibrium: $k \frac{q_A q_B}{a^2} \hat{r}_{AB} + k \frac{q_A q_C}{a^2} \hat{r}_{AC} = 0$. Assuming $q_A \neq 0$, we need $q_B \hat{r}_{AB} + q_C \hat{r}_{AC} = 0$. Resolving this vector equation into components (with the angle between $\hat{r}_{AB}$ and $\hat{r}_{AC}$ being $60^\circ$) leads to the conditions:

1. $q_B = q_C$ (component perpendicular to the line connecting $q_B$ and $q_C$)
2. $q_B + q_C = 0$ (component along the line connecting $q_B$ and $q_C$, assuming appropriate directions)

These two conditions ($q_B = q_C$ and $q_B = -q_C$) can only be satisfied if $q_B = q_C = 0$. If $q_B = q_C = 0$, then the forces on $q_B$ and $q_C$ are also zero, meaning they are in equilibrium. Thus, a system with charges $(q_A, 0, 0)$ is in equilibrium. Since an equilibrium configuration exists, Statement I ("the system can never be in equilibrium") is false.

Analysis of Statement II:

Consider three point charges $q_1, q_2, q_3$ lying on a straight line at positions $x_1, x_2, x_3$. Let $d_1 = x_2 - x_1$ and $d_2 = x_3 - x_2$. Assume $d_1 > 0$ and $d_2 > 0$.

For the middle charge $q_2$ to be in equilibrium: $k \frac{q_2 q_1}{d_1^2} + k \frac{q_2 q_3}{d_2^2} = 0$. If $q_2 \neq 0$, then $\frac{q_1}{d_1^2} + \frac{q_3}{d_2^2} = 0$. This implies $q_1$ and $q_3$ must have opposite signs.

For the charge $q_1$ to be in equilibrium: $k \frac{q_1 q_2}{d_1^2} + k \frac{q_1 q_3}{(d_1+d_2)^2} = 0$. If $q_1 \neq 0$, then $\frac{q_2}{d_1^2} + \frac{q_3}{(d_1+d_2)^2} = 0 \implies q_2 = -q_3 \frac{d_1^2}{(d_1+d_2)^2}$.

For the charge $q_3$ to be in equilibrium: $k \frac{q_3 q_1}{(d_1+d_2)^2} + k \frac{q_3 q_2}{d_2^2} = 0$. If $q_3 \neq 0$, then $\frac{q_1}{(d_1+d_2)^2} + \frac{q_2}{d_2^2} = 0 \implies q_2 = -q_1 \frac{d_2^2}{(d_1+d_2)^2}$.

Equating the two expressions for $q_2$: $-q_3 \frac{d_1^2}{(d_1+d_2)^2} = -q_1 \frac{d_2^2}{(d_1+d_2)^2}$ $q_3 d_1^2 = q_1 d_2^2$.

From the equilibrium of $q_2$, we know $q_1$ and $q_3$ must have opposite signs. However, the equation $q_3 d_1^2 = q_1 d_2^2$ implies that $q_1$ and $q_3$ must have the same sign (since $d_1^2$ and $d_2^2$ are positive). This is a contradiction. Therefore, it is impossible for three non-zero point charges lying on a straight line to be in equilibrium simultaneously. Statement II ("the system of three point charges lying on a straight line can be in equilibrium") is false.

Conclusion: Both Statement I and Statement II are false.