Two point charges `100 mu C` and `5 mu C` are placed at points `A and B` respectively with `AB = 40 cm`. The work done by external froce in displacing the charge `5 mu C` from B to C, where `BC = 30 cm`, angle `ABC = (pi)/(2)` and `(1)/(4pi epsilon_(0)) = 9 xx 10^(9)Nm^(2)//C^(2)`

Options

A. `9 J`
B. `(81)/(20) J`
C. `(9)/(25) J`
D. `-(9)/(4) J`

Answer

To solve the problem step by step, we will calculate the work done by the external force in displacing the charge \(5 \mu C\) from point B to point C.

### Step 1: Understand the Geometry
We have two point charges:
- Charge \(Q_1 = 100 \mu C\) at point A
- Charge \(Q_2 = 5 \mu C\) at point B

The distance \(AB = 40 cm\) and the distance \(BC = 30 cm\) with angle \(ABC = \frac{\pi}{2}\).

### Step 2: Calculate the Distance AC
Using the Pythagorean theorem:
\[ AC = \sqrt{AB^2 + BC^2} \]
Substituting the values:
\[ AC = \sqrt{(40 \, cm)^2 + (30 \, cm)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, cm \]

### Step 3: Calculate Initial Potential Energy (U_initial)
The initial potential energy \(U_i\) when the charge \(5 \mu C\) is at point B (distance \(R = 40 cm\)):
\[ U_i = k \frac{Q_1 Q_2}{R} \]
Where \(k = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\), \(Q_1 = 100 \mu C = 100 \times 10^{-6} C\), \(Q_2 = 5 \mu C = 5 \times 10^{-6} C\), and \(R = 40 \, cm = 0.4 \, m\):
\[ U_i = 9 \times 10^9 \cdot \frac{(100 \times 10^{-6})(5 \times 10^{-6})}{0.4} \]
Calculating:
\[ U_i = 9 \times 10^9 \cdot \frac{500 \times 10^{-12}}{0.4} = 9 \times 10^9 \cdot 1250 \times 10^{-12} = 11.25 \, J \]

### Step 4: Calculate Final Potential Energy (U_final)
The final potential energy \(U_f\) when the charge \(5 \mu C\) is at point C (distance \(R' = 50 cm\)):
\[ U_f = k \frac{Q_1 Q_2}{R'} \]
Where \(R' = 50 \, cm = 0.5 \, m\):
\[ U_f = 9 \times 10^9 \cdot \frac{(100 \times 10^{-6})(5 \times 10^{-6})}{0.5} \]
Calculating:
\[ U_f = 9 \times 10^9 \cdot \frac{500 \times 10^{-12}}{0.5} = 9 \times 10^9 \cdot 1000 \times 10^{-12} = 9 \, J \]

### Step 5: Calculate Change in Potential Energy (ΔU)
\[ \Delta U = U_f - U_i = 9 \, J - 11.25 \, J = -2.25 \, J \]

### Step 6: Work Done by External Force
The work done by the external force is equal to the change in potential energy:
\[ W = \Delta U = -2.25 \, J \]

### Final Answer
The work done by the external force in displacing the charge \(5 \mu C\) from B to C is \(-2.25 \, J\).

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