Answer :
Certainly! Let's solve this step-by-step:
### Given Data:
1. Width of the rectangle [tex]\( \text{width} = 0.15 \, \text{m} \)[/tex]
2. Height of the rectangle [tex]\( \text{height} = 0.05 \, \text{m} \)[/tex]
3. Charge [tex]\( q_1 = -5 \times 10^{-6} \, \text{C} \)[/tex]
4. Charge [tex]\( q_2 = 2 \times 10^{-6} \, \text{C} \)[/tex]
5. Coulomb's constant [tex]\( k = 8.988 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \)[/tex]
### Objective:
Determine the electric potential at the upper right-hand corner of the rectangle.
### Steps to Solve:
1. Distance Calculation:
- The distance from charge [tex]\( q_1 \)[/tex] (located at the upper left-hand corner) to the upper right-hand corner is equal to the width of the rectangle.
[tex]\[ d_1 = 0.15 \, \text{m} \][/tex]
- The distance from charge [tex]\( q_2 \)[/tex] (located at the lower right-hand corner) to the upper right-hand corner can be calculated using the Pythagorean theorem:
[tex]\[ d_2 = \sqrt{(\text{width})^2 + (\text{height})^2} = \sqrt{(0.15)^2 + (0.05)^2} = 0.15811388300841897 \, \text{m} \][/tex]
2. Electric Potential Calculation:
- The electric potential at a point due to a charge is given by:
[tex]\[ V = \frac{k \cdot q}{d} \][/tex]
- Calculate the electric potential at the upper right corner due to [tex]\( q_1 \)[/tex]:
[tex]\[ V_1 = \frac{8.988 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \times (-5 \times 10^{-6} \, \text{C})}{0.15 \, \text{m}} = -299599.99999999994 \, \text{V} \][/tex]
- Calculate the electric potential at the upper right corner due to [tex]\( q_2 \)[/tex]:
[tex]\[ V_2 = \frac{8.988 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \times 2 \times 10^{-6} \, \text{C}}{0.15811388300841897 \, \text{m}} = 113690.20643837357 \, \text{V} \][/tex]
3. Total Electric Potential Calculation:
- The total electric potential at the upper right corner is the algebraic sum of the potentials due to both charges:
[tex]\[ V_{\text{total}} = V_1 + V_2 = -299599.99999999994 \, \text{V} + 113690.20643837357 \, \text{V} = -185909.79356162637 \, \text{V} \][/tex]
### Conclusion:
The electric potential at the upper right-hand corner of the rectangle is [tex]\(-185909.79356162637 \, \text{V}\)[/tex].
### Given Data:
1. Width of the rectangle [tex]\( \text{width} = 0.15 \, \text{m} \)[/tex]
2. Height of the rectangle [tex]\( \text{height} = 0.05 \, \text{m} \)[/tex]
3. Charge [tex]\( q_1 = -5 \times 10^{-6} \, \text{C} \)[/tex]
4. Charge [tex]\( q_2 = 2 \times 10^{-6} \, \text{C} \)[/tex]
5. Coulomb's constant [tex]\( k = 8.988 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \)[/tex]
### Objective:
Determine the electric potential at the upper right-hand corner of the rectangle.
### Steps to Solve:
1. Distance Calculation:
- The distance from charge [tex]\( q_1 \)[/tex] (located at the upper left-hand corner) to the upper right-hand corner is equal to the width of the rectangle.
[tex]\[ d_1 = 0.15 \, \text{m} \][/tex]
- The distance from charge [tex]\( q_2 \)[/tex] (located at the lower right-hand corner) to the upper right-hand corner can be calculated using the Pythagorean theorem:
[tex]\[ d_2 = \sqrt{(\text{width})^2 + (\text{height})^2} = \sqrt{(0.15)^2 + (0.05)^2} = 0.15811388300841897 \, \text{m} \][/tex]
2. Electric Potential Calculation:
- The electric potential at a point due to a charge is given by:
[tex]\[ V = \frac{k \cdot q}{d} \][/tex]
- Calculate the electric potential at the upper right corner due to [tex]\( q_1 \)[/tex]:
[tex]\[ V_1 = \frac{8.988 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \times (-5 \times 10^{-6} \, \text{C})}{0.15 \, \text{m}} = -299599.99999999994 \, \text{V} \][/tex]
- Calculate the electric potential at the upper right corner due to [tex]\( q_2 \)[/tex]:
[tex]\[ V_2 = \frac{8.988 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \times 2 \times 10^{-6} \, \text{C}}{0.15811388300841897 \, \text{m}} = 113690.20643837357 \, \text{V} \][/tex]
3. Total Electric Potential Calculation:
- The total electric potential at the upper right corner is the algebraic sum of the potentials due to both charges:
[tex]\[ V_{\text{total}} = V_1 + V_2 = -299599.99999999994 \, \text{V} + 113690.20643837357 \, \text{V} = -185909.79356162637 \, \text{V} \][/tex]
### Conclusion:
The electric potential at the upper right-hand corner of the rectangle is [tex]\(-185909.79356162637 \, \text{V}\)[/tex].