Question

Figure given shows four arrangement of charged particles, all at the same distance from the origin. Rank the situations according to the net electric potentials (V1, V2, V3, V4) at the origin, most positive first

• A. V1 > V2 > V3 > V4
• B. V2 > V1 > V3 > V4
• C. V2 > V1 > V4 > V3
• D. V4 > V1 > V3 > V2

Answer 1

Answer: (B)

V2 > V1 > V3 > V4

Answer 2

The electric potential $V$ at the origin due to a system of point charges is the algebraic sum of the potentials due to individual charges. Since all charges are at the same distance $r$ from the origin, the potential due to each charge $q_i$ is $V_i = \frac{kq_i}{r}$.
The net potential at the origin is $V_{net} = \sum_{i} V_i = \frac{k}{r} \sum_{i} q_i$.
We need to calculate the sum of charges for each arrangement. Let $q$ be a positive quantity.

1. Arrangement 1 (V1):
   Charges are: +q, +q, -q, -q
   Sum of charges = (+q) + (+q) + (-q) + (-q) = 2q - 2q = 0
   Therefore, $V_1 = \frac{k}{r} (0) = 0$.

2. Arrangement 2 (V2):
   Charges are: +q, +q, +q, -q
   Sum of charges = (+q) + (+q) + (+q) + (-q) = 3q - q = 2q
   Therefore, $V_2 = \frac{k}{r} (2q) = \frac{2kq}{r}$.

3. Arrangement 3 (V3):
   Charges are: +q, -q, -q, -q
   Sum of charges = (+q) + (-q) + (-q) + (-q) = q - 3q = -2q
   Therefore, $V_3 = \frac{k}{r} (-2q) = -\frac{2kq}{r}$.

4. Arrangement 4 (V4):
   Charges are: +q, +q, -q, +q
   Sum of charges = (+q) + (+q) + (-q) + (+q) = 3q - q = 2q
   Therefore, $V_4 = \frac{k}{r} (2q) = \frac{2kq}{r}$.

Now, let's compare the calculated potentials:
$V_1 = 0$
$V_2 = \frac{2kq}{r}$
$V_3 = -\frac{2kq}{r}$
$V_4 = \frac{2kq}{r}$

Assuming $q$ is a positive charge (as is standard unless specified otherwise), and $k$ and $r$ are positive constants:
We observe that $V_2 = V_4$.
The ranking from most positive first is:
$V_2 = V_4 > V_1 > V_3$

The final answer is $\boxed{V2 > V1 > V3 > V4}$