To solve the problem, we need to determine the charge \( Q \) placed at point \( A(x_0, y_0) \) and the coordinates \( (x_0, y_0) \) based on the given electric field and potential at point \( B(9\,m, 4\,m) \).
### Step 1: Calculate the Magnitude of the Electric Field
The electric field \( \vec{E} \) at point \( B \) is given as:
\[
\vec{E} = 54 \hat{i} + 72 \hat{j} \, \text{N/C}
\]
To find the magnitude of the electric field:
\[
|\vec{E}| = \sqrt{(54)^2 + (72)^2}
\]
Calculating this:
\[
|\vec{E}| = \sqrt{2916 + 5184} = \sqrt{8100} = 90 \, \text{N/C}
\]
### Step 2: Relate Electric Field to Charge and Distance
The electric field \( E \) due to a point charge \( Q \) at a distance \( R \) is given by:
\[
E = \frac{kQ}{R^2}
\]
Where \( k \) is Coulomb's constant \( (k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2) \).
### Step 3: Calculate the Potential at Point B
The potential \( V \) at point \( B \) is given as:
\[
V = 1800 \, \text{V}
\]
The potential due to a point charge is given by:
\[
V = \frac{kQ}{R}
\]
### Step 4: Set Up the Equations
From the electric field equation:
\[
90 = \frac{9 \times 10^9 \cdot Q}{R^2} \quad \text{(1)}
\]
From the potential equation:
\[
1800 = \frac{9 \times 10^9 \cdot Q}{R} \quad \text{(2)}
\]
### Step 5: Solve for R
Dividing equation (2) by equation (1):
\[
\frac{1800}{90} = \frac{R}{R^2} \implies 20 = \frac{R}{Q} \implies R = 20Q
\]
### Step 6: Substitute R in Equation (1)
Substituting \( R = 20 \) into equation (1):
\[
90 = \frac{9 \times 10^9 \cdot Q}{(20)^2}
\]
\[
90 = \frac{9 \times 10^9 \cdot Q}{400}
\]
\[
Q = \frac{90 \cdot 400}{9 \times 10^9} = \frac{36000}{9 \times 10^9} = 4 \times 10^{-6} \, \text{C}
\]
Thus, \( Q = 4 \, \mu C \).
### Step 7: Determine the Coordinates \( (x_0, y_0) \)
The distance \( R \) between points \( A(x_0, y_0) \) and \( B(9, 4) \) is given by:
\[
R = \sqrt{(9 - x_0)^2 + (4 - y_0)^2}
\]
We know \( R = 20 \):
\[
20 = \sqrt{(9 - x_0)^2 + (4 - y_0)^2}
\]
Squaring both sides:
\[
400 = (9 - x_0)^2 + (4 - y_0)^2 \quad \text{(3)}
\]
### Step 8: Find the Unit Vector Direction of Electric Field
The unit vector in the direction of the electric field can be expressed as:
\[
\hat{e} = \frac{54 \hat{i} + 72 \hat{j}}{90}
\]
This simplifies to:
\[
\hat{e} = \frac{2}{9} \hat{i} + \frac{8}{9} \hat{j}
\]
### Step 9: Set Up the Equations for Coordinates
From the unit vector, we can set:
\[
9 - x_0 = 12 \quad \text{(from } \frac{2}{9} \times 90 = 20\text{)}
\]
\[
4 - y_0 = 16 \quad \text{(from } \frac{8}{9} \times 90 = 80\text{)}
\]
Solving these gives:
\[
x_0 = 9 - 12 = -3
\]
\[
y_0 = 4 - 16 = -12
\]
### Final Answer
Thus, the charge \( Q = 4 \, \mu C \) and the coordinates \( (x_0, y_0) = (-3, -12) \).
