Given are four arrangements of three fixed electric charges. In each arrangement, a point labeled P is also identified. A test charge `+q` is placed at P. All of the charges are of the same magnitude Q, but they can be either positive or negative as indicated. The charged and point P all lie on a straight line. the distances between adjacent items, either between two charges or between a charge and point P, are all the same.
I. `o+,o+,o+,underset (P)(*)`, II. `o+,o+,overset(P)(*),-`
III. `o+,o+,-, overset(P)(*)`, IV. `o+, - , o+, overset(P)(*)`.
Correct order of choices in a decreasing order of magnitude of force on P is

Options

A. `IIgtIgtIIIgtIV`
B. `IgtIIgtIIIgtIV`
C. `IIgtIgtIVgtIII`
D. `IIIgtIVgtIgtII`

Explanation

To solve the problem, we need to analyze the forces acting on the test charge \( +q \) at point P for each of the four arrangements of charges. We will calculate the net force on the test charge for each arrangement and then compare them.

### Step 1: Analyze Arrangement I
**Arrangement I:** \( +Q, +Q, +Q, P \)

- The forces acting on \( +q \) at point P due to the three \( +Q \) charges are all repulsive.
- The distances from \( P \) to each charge are equal, say \( r \).
- The force due to each charge is given by Coulomb's law:
\[ F = k \frac{Qq}{r^2} \]
- Since all three forces are repulsive and equal, the total force \( F_1 \) on \( P \) is:
\[ F_1 = 3 \cdot k \frac{Qq}{r^2} \]

### Step 2: Analyze Arrangement II
**Arrangement II:** \( +Q, +Q, P, -Q \)

- The force from the first \( +Q \) charge on \( P \) is attractive:
\[ F_{1} = k \frac{Qq}{r^2} \]
- The force from the second \( +Q \) charge on \( P \) is also attractive:
\[ F_{2} = k \frac{Qq}{r^2} \]
- The force from the \( -Q \) charge on \( P \) is repulsive:
\[ F_{3} = k \frac{Qq}{r^2} \]
- The net force \( F_2 \) on \( P \) is:
\[ F_2 = F_1 + F_2 - F_3 = k \frac{Qq}{r^2} + k \frac{Qq}{r^2} - k \frac{Qq}{r^2} = k \frac{Qq}{r^2} \]

### Step 3: Analyze Arrangement III
**Arrangement III:** \( +Q, +Q, -Q, P \)

- The force from the first \( +Q \) charge on \( P \) is attractive:
\[ F_{1} = k \frac{Qq}{r^2} \]
- The force from the second \( +Q \) charge on \( P \) is also attractive:
\[ F_{2} = k \frac{Qq}{r^2} \]
- The force from the \( -Q \) charge on \( P \) is repulsive:
\[ F_{3} = k \frac{Qq}{(2r)^2} = \frac{k Qq}{4r^2} \]
- The net force \( F_3 \) on \( P \) is:
\[ F_3 = F_1 + F_2 - F_3 = k \frac{Qq}{r^2} + k \frac{Qq}{r^2} - \frac{k Qq}{4r^2} = \frac{4k Qq}{4r^2} - \frac{k Qq}{4r^2} = \frac{3k Qq}{4r^2} \]

### Step 4: Analyze Arrangement IV
**Arrangement IV:** \( +Q, -Q, +Q, P \)

- The force from the first \( +Q \) charge on \( P \) is repulsive:
\[ F_{1} = k \frac{Qq}{r^2} \]
- The force from the \( -Q \) charge on \( P \) is attractive:
\[ F_{2} = k \frac{Qq}{r^2} \]
- The force from the last \( +Q \) charge on \( P \) is also repulsive:
\[ F_{3} = k \frac{Qq}{(2r)^2} = \frac{k Qq}{4r^2} \]
- The net force \( F_4 \) on \( P \) is:
\[ F_4 = F_1 - F_2 + F_3 = k \frac{Qq}{r^2} - k \frac{Qq}{r^2} + \frac{k Qq}{4r^2} = \frac{k Qq}{4r^2} \]

### Step 5: Compare the Forces
Now we have the following forces for each arrangement:
1. \( F_1 = 3k \frac{Qq}{r^2} \)
2. \( F_2 = k \frac{Qq}{r^2} \)
3. \( F_3 = \frac{3k Qq}{4r^2} \)
4. \( F_4 = \frac{k Qq}{4r^2} \)

### Step 6: Order the Forces
Now we can order the forces in decreasing magnitude:
- \( F_1 > F_3 > F_2 > F_4 \)

### Conclusion
The correct order of choices in decreasing order of magnitude of force on \( P \) is:
1. Arrangement I
2. Arrangement III
3. Arrangement II
4. Arrangement IV