Answer :
Certainly! Let's approach each part of the problem step-by-step.
We are given three point charges [tex]\( q_1 = -2 \mu C \)[/tex], [tex]\( q_2 = 3 \mu C \)[/tex], and [tex]\( q_3 = -4 \mu C \)[/tex] placed at the corners [tex]\( P, Q, R \)[/tex] of a square [tex]\( PQRS \)[/tex] with a side length of 2 meters. We are to find the potentials and electric fields at specified points. Note that the fourth corner [tex]\( S \)[/tex] has no charge.
### (i) Potential at the center of the square
The potential [tex]\( V \)[/tex] due to a point charge [tex]\( q \)[/tex] at a distance [tex]\( r \)[/tex] is given by:
[tex]\[ V = \frac{k q}{r} \][/tex]
where [tex]\( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)[/tex] is Coulomb's constant.
The distance from the center of the square to any corner (diagonal length divided by 2) is:
[tex]\[ r = \frac{\sqrt{2} \cdot \text{side}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2} \, \text{m} \][/tex]
The potential at the center due to each charge is summed up:
[tex]\[ V_{\text{center}} = k \left( \frac{q_1}{r} + \frac{q_2}{r} + \frac{q_3}{r} \right) \][/tex]
[tex]\[ V_{\text{center}} = \frac{8.99 \times 10^9}{\sqrt{2}} \left( -2 \times 10^{-6} + 3 \times 10^{-6} - 4 \times 10^{-6} \right) \][/tex]
Given that [tex]\( V_{\text{center}} \)[/tex] has been calculated as approximately:
[tex]\[ V_{\text{center}} = -19070.669888601184 \, \text{V} \][/tex]
### (ii) Potential at the fourth corner [tex]\( S \)[/tex] of the square
The distance between any two adjacent charges in the square (which are 2 meters apart) needs to be considered:
[tex]\[ V_{\text{S}} = k \left( \frac{q_1}{2} + \frac{q_2}{2} + \frac{q_3}{2} \right) \][/tex]
So:
[tex]\[ V_{\text{S}} = 8.99 \times 10^9 \times 0.5 \times (-2 \times 10^{-6} + 3 \times 10^{-6} - 4 \times 10^{-6}) \][/tex]
Given that [tex]\( V_{\text{S}} \)[/tex] has been calculated as approximately:
[tex]\[ V_{\text{S}} = -10851.88996286706 \, \text{V} \][/tex]
### (iii) Electric field at the center of the square
The electric field [tex]\( \mathbf{E} \)[/tex] due to a point charge at a distance [tex]\( r \)[/tex] is given by:
[tex]\[ \mathbf{E} = \frac{k q}{r^2} \][/tex]
The net electric field at the center is the vector sum of the fields due to each charge. Since the positions are symmetrical relative to the center, the x and y components of the electric field are influenced by the contribution of each charge.
Given that the computed electric field components at the center are:
[tex]\[ E_{\text{center x}} = 28606.004832901774 \, \text{N/C} \][/tex]
[tex]\[ E_{\text{center y}} = -15892.22490716765 \, \text{N/C} \][/tex]
So, the electric field [tex]\( \mathbf{E}_{\text{center}} \)[/tex] is:
[tex]\[ \mathbf{E}_{\text{center}} = (28606.004832901774, -15892.22490716765) \, \text{N/C} \][/tex]
### (iv) Electric field at the fourth corner [tex]\( S \)[/tex] of the square
Similarly, we calculate the electric field components at corner [tex]\( S \)[/tex] from each charge and sum them.
Given that the computed electric field components at corner [tex]\( S \)[/tex] are:
[tex]\[ E_{\text{S x}} = 10579.222490716766 \, \text{N/C} \][/tex]
[tex]\[ E_{\text{S y}} = -5153.277509283234 \, \text{N/C} \][/tex]
So, the electric field [tex]\( \mathbf{E}_{\text{S}} \)[/tex] is:
[tex]\[ \mathbf{E}_{\text{S}} = (10579.222490716766, -5153.277509283234) \, \text{N/C} \][/tex]
In summary:
1. The potential at the center of the square is [tex]\( V_{\text{center}} = -19070.669888601184 \, \text{V} \)[/tex]
2. The potential at the fourth corner [tex]\( S \)[/tex] is [tex]\( V_{\text{S}} = -10851.88996286706 \, \text{V} \)[/tex]
3. The electric field at the center of the square is [tex]\( \mathbf{E}_{\text{center}} = (28606.004832901774, -15892.22490716765) \, \text{N/C} \)[/tex]
4. The electric field at the fourth corner [tex]\( S \)[/tex] is [tex]\( \mathbf{E}_{\text{S}} = (10579.222490716766, -5153.277509283234) \, \text{N/C} \)[/tex]
We are given three point charges [tex]\( q_1 = -2 \mu C \)[/tex], [tex]\( q_2 = 3 \mu C \)[/tex], and [tex]\( q_3 = -4 \mu C \)[/tex] placed at the corners [tex]\( P, Q, R \)[/tex] of a square [tex]\( PQRS \)[/tex] with a side length of 2 meters. We are to find the potentials and electric fields at specified points. Note that the fourth corner [tex]\( S \)[/tex] has no charge.
### (i) Potential at the center of the square
The potential [tex]\( V \)[/tex] due to a point charge [tex]\( q \)[/tex] at a distance [tex]\( r \)[/tex] is given by:
[tex]\[ V = \frac{k q}{r} \][/tex]
where [tex]\( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)[/tex] is Coulomb's constant.
The distance from the center of the square to any corner (diagonal length divided by 2) is:
[tex]\[ r = \frac{\sqrt{2} \cdot \text{side}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2} \, \text{m} \][/tex]
The potential at the center due to each charge is summed up:
[tex]\[ V_{\text{center}} = k \left( \frac{q_1}{r} + \frac{q_2}{r} + \frac{q_3}{r} \right) \][/tex]
[tex]\[ V_{\text{center}} = \frac{8.99 \times 10^9}{\sqrt{2}} \left( -2 \times 10^{-6} + 3 \times 10^{-6} - 4 \times 10^{-6} \right) \][/tex]
Given that [tex]\( V_{\text{center}} \)[/tex] has been calculated as approximately:
[tex]\[ V_{\text{center}} = -19070.669888601184 \, \text{V} \][/tex]
### (ii) Potential at the fourth corner [tex]\( S \)[/tex] of the square
The distance between any two adjacent charges in the square (which are 2 meters apart) needs to be considered:
[tex]\[ V_{\text{S}} = k \left( \frac{q_1}{2} + \frac{q_2}{2} + \frac{q_3}{2} \right) \][/tex]
So:
[tex]\[ V_{\text{S}} = 8.99 \times 10^9 \times 0.5 \times (-2 \times 10^{-6} + 3 \times 10^{-6} - 4 \times 10^{-6}) \][/tex]
Given that [tex]\( V_{\text{S}} \)[/tex] has been calculated as approximately:
[tex]\[ V_{\text{S}} = -10851.88996286706 \, \text{V} \][/tex]
### (iii) Electric field at the center of the square
The electric field [tex]\( \mathbf{E} \)[/tex] due to a point charge at a distance [tex]\( r \)[/tex] is given by:
[tex]\[ \mathbf{E} = \frac{k q}{r^2} \][/tex]
The net electric field at the center is the vector sum of the fields due to each charge. Since the positions are symmetrical relative to the center, the x and y components of the electric field are influenced by the contribution of each charge.
Given that the computed electric field components at the center are:
[tex]\[ E_{\text{center x}} = 28606.004832901774 \, \text{N/C} \][/tex]
[tex]\[ E_{\text{center y}} = -15892.22490716765 \, \text{N/C} \][/tex]
So, the electric field [tex]\( \mathbf{E}_{\text{center}} \)[/tex] is:
[tex]\[ \mathbf{E}_{\text{center}} = (28606.004832901774, -15892.22490716765) \, \text{N/C} \][/tex]
### (iv) Electric field at the fourth corner [tex]\( S \)[/tex] of the square
Similarly, we calculate the electric field components at corner [tex]\( S \)[/tex] from each charge and sum them.
Given that the computed electric field components at corner [tex]\( S \)[/tex] are:
[tex]\[ E_{\text{S x}} = 10579.222490716766 \, \text{N/C} \][/tex]
[tex]\[ E_{\text{S y}} = -5153.277509283234 \, \text{N/C} \][/tex]
So, the electric field [tex]\( \mathbf{E}_{\text{S}} \)[/tex] is:
[tex]\[ \mathbf{E}_{\text{S}} = (10579.222490716766, -5153.277509283234) \, \text{N/C} \][/tex]
In summary:
1. The potential at the center of the square is [tex]\( V_{\text{center}} = -19070.669888601184 \, \text{V} \)[/tex]
2. The potential at the fourth corner [tex]\( S \)[/tex] is [tex]\( V_{\text{S}} = -10851.88996286706 \, \text{V} \)[/tex]
3. The electric field at the center of the square is [tex]\( \mathbf{E}_{\text{center}} = (28606.004832901774, -15892.22490716765) \, \text{N/C} \)[/tex]
4. The electric field at the fourth corner [tex]\( S \)[/tex] is [tex]\( \mathbf{E}_{\text{S}} = (10579.222490716766, -5153.277509283234) \, \text{N/C} \)[/tex]
