A positive charge `Q=50 muC` is located in the xy plane at a point having position vector `vec(r)_(0)=(2hat(i)+3hat(j))m`
where `hat(i)` and `hat(j)` are unit vectors in the positive directions of X and Y axis respectively. Find :
(a) The electric intensity vector and its magnitude at a point having co-ordinates (8m, -5m).
(b) Work done by external agent in transporting a charge `q=10 muC` from (8m, 6m) to the point (4m, 3m).

Answer

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (a): Finding the Electric Intensity Vector and Its Magnitude

1. **Identify Given Values**:
- Charge \( Q = 50 \, \mu C = 50 \times 10^{-6} \, C \)
- Position of charge \( \vec{r}_0 = 2\hat{i} + 3\hat{j} \, m \)
- Point of interest \( \vec{r} = 8\hat{i} - 5\hat{j} \, m \)

2. **Calculate the Displacement Vector**:
- The displacement vector \( \vec{R} \) from the charge to the point of interest is given by:
\[ \vec{R} = \vec{r} - \vec{r}_0 = (8\hat{i} - 5\hat{j}) - (2\hat{i} + 3\hat{j}) = (8 - 2)\hat{i} + (-5 - 3)\hat{j} = 6\hat{i} - 8\hat{j} \, m \]

3. **Calculate the Magnitude of the Displacement Vector**:
- The magnitude \( R \) of the vector \( \vec{R} \) is calculated as:
\[ R = |\vec{R}| = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, m \]

4. **Calculate the Electric Field Intensity Vector**:
- The electric field \( \vec{E} \) due to a point charge is given by:
\[ \vec{E} = \frac{1}{4\pi \epsilon_0} \frac{Q}{R^2} \hat{R} \]
- Here, \( \hat{R} \) is the unit vector in the direction of \( \vec{R} \):
\[ \hat{R} = \frac{\vec{R}}{|\vec{R}|} = \frac{6\hat{i} - 8\hat{j}}{10} = 0.6\hat{i} - 0.8\hat{j} \]
- Substituting \( K = \frac{1}{4\pi \epsilon_0} \approx 9 \times 10^9 \, N m^2/C^2 \):
\[ \vec{E} = 9 \times 10^9 \cdot \frac{50 \times 10^{-6}}{10^2} (0.6\hat{i} - 0.8\hat{j}) \]
\[ \vec{E} = 9 \times 10^9 \cdot \frac{50 \times 10^{-6}}{100} (0.6\hat{i} - 0.8\hat{j}) \]
\[ \vec{E} = 9 \times 10^9 \cdot 0.5 \times 10^{-6} (0.6\hat{i} - 0.8\hat{j}) \]
\[ \vec{E} = 4.5 \times 10^3 (0.6\hat{i} - 0.8\hat{j}) \, N/C \]
\[ \vec{E} = 2700\hat{i} - 3600\hat{j} \, N/C \]

5. **Calculate the Magnitude of the Electric Field**:
- The magnitude of \( \vec{E} \) is:
\[ |\vec{E}| = \sqrt{(2700)^2 + (-3600)^2} = \sqrt{7290000 + 12960000} = \sqrt{20250000} \approx 4500 \, N/C \]

### Part (b): Work Done by External Agent

1. **Identify Initial and Final Positions**:
- Initial position \( \vec{r}_1 = 8\hat{i} + 6\hat{j} \)
- Final position \( \vec{r}_2 = 4\hat{i} + 3\hat{j} \)

2. **Calculate Distances from Charge to Initial and Final Positions**:
- For \( \vec{r}_1 \):
\[ R_1 = |\vec{r}_1 - \vec{r}_0| = |(8\hat{i} + 6\hat{j}) - (2\hat{i} + 3\hat{j})| = |(6\hat{i} + 3\hat{j})| = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \, m \]
- For \( \vec{r}_2 \):
\[ R_2 = |\vec{r}_2 - \vec{r}_0| = |(4\hat{i} + 3\hat{j}) - (2\hat{i} + 3\hat{j})| = |(2\hat{i})| = 2 \, m \]

3. **Calculate Work Done**:
- The work done \( W \) by an external agent is given by:
\[ W = q(V_1 - V_2) \]
- The potential \( V \) at a distance \( R \) from charge \( Q \) is given by:
\[ V = \frac{KQ}{R} \]
- Therefore, we can write:
\[ W = q \left( \frac{KQ}{R_1} - \frac{KQ}{R_2} \right) \]
- Substituting \( q = 10 \, \mu C = 10 \times 10^{-6} \, C \), \( K = 9 \times 10^9 \), and \( Q = 50 \times 10^{-6} \):
\[ W = 10 \times 10^{-6} \left( \frac{9 \times 10^9 \times 50 \times 10^{-6}}{3\sqrt{5}} - \frac{9 \times 10^9 \times 50 \times 10^{-6}}{2} \right) \]
\[ W = 10 \times 10^{-6} \cdot 9 \times 10^9 \times 50 \times 10^{-6} \left( \frac{1}{3\sqrt{5}} - \frac{1}{2} \right) \]
- Evaluating the expression gives:
\[ W \approx 0.45 \, J \]

### Final Answers:
- (a) Electric intensity vector: \( \vec{E} = 2700\hat{i} - 3600\hat{j} \, N/C \) with magnitude \( 4500 \, N/C \)
- (b) Work done by external agent: \( W \approx 0.45 \, J \)