Answer :
Certainly! Let's work through the steps necessary to determine the electric field at a point 5.0 cm from the negative charge, [tex]$q_2$[/tex], and along the line between the two charges.
Given:
- [tex]\( q_1 = 2.0 \times 10^{-7} \, C \)[/tex]
- [tex]\( q_2 = -6.0 \times 10^{-8} \, C \)[/tex]
- Distance between [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] is 25.0 cm (0.25 m)
- We need the electric field at a point that is 5.0 cm (0.05 m) from [tex]\( q_2 \)[/tex].
First, we need to determine the electric field contributions from both charges at the specified point.
1. Distance from the point to [tex]\( q_1 \)[/tex]:
Since the total distance between the charges is 0.25 m and the point is 0.05 m from [tex]\( q_2 \)[/tex],
[tex]\[ \text{Distance from point to } q_1 = 0.25 \, \text{m} - 0.05 \, \text{m} = 0.20 \, \text{m}. \][/tex]
2. Coulomb's Law:
The electric field [tex]\( E \)[/tex] created by a point charge [tex]\( q \)[/tex] at a distance [tex]\( r \)[/tex] is given by:
[tex]\[ E = \frac{k_e \cdot |q|}{r^2}, \][/tex]
where [tex]\( k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 \cdot \text{C}^{-2} \)[/tex].
3. Electric Field due to [tex]\( q_1 \)[/tex]:
[tex]\[ E_1 = \frac{(8.99 \times 10^9) \cdot (2.0 \times 10^{-7})}{(0.20)^2} = 44949.99999999999 \, \text{N/C}. \][/tex]
The direction of the electric field due to a positive charge is away from the charge.
4. Electric Field due to [tex]\( q_2 \)[/tex]:
[tex]\[ E_2 = \frac{(8.99 \times 10^9) \cdot (6.0 \times 10^{-8})}{(0.05)^2} = 215759.99999999994 \, \text{N/C}. \][/tex]
Since [tex]\( q_2 \)[/tex] is negative, the electric field direction is towards the charge.
5. Net Electric Field:
Both the electric fields [tex]\( E_1 \)[/tex] and [tex]\( E_2 \)[/tex] are along the same line between the charges. The total electric field at the point will be the vector sum of these fields. Since [tex]\( E_1 \)[/tex] is directed away from [tex]\( q_1 \)[/tex] and [tex]\( E_2 \)[/tex] is directed towards [tex]\( q_2 \)[/tex], they are in opposite directions.
Therefore, we can subtract the magnitudes to find the net electric field:
[tex]\[ E_{\text{net}} = E_1 - E_2 = 44949.99999999999 - 215759.99999999994 = -170809.99999999994 \, \text{N/C}. \][/tex]
The negative sign in the net electric field indicates that the direction of the net electric field is towards [tex]\( q_2 \)[/tex].
In summary:
- The electric field due to [tex]\( q_1 \)[/tex] is approximately [tex]\( 44950 \, \text{N/C} \)[/tex] away from [tex]\( q_1 \)[/tex].
- The electric field due to [tex]\( q_2 \)[/tex] is approximately [tex]\( 215760 \, \text{N/C} \)[/tex] towards [tex]\( q_2 \)[/tex].
- The net electric field at the point 5.0 cm from the negative charge is approximately [tex]\( -170810 \, \text{N/C} \)[/tex].
Given:
- [tex]\( q_1 = 2.0 \times 10^{-7} \, C \)[/tex]
- [tex]\( q_2 = -6.0 \times 10^{-8} \, C \)[/tex]
- Distance between [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] is 25.0 cm (0.25 m)
- We need the electric field at a point that is 5.0 cm (0.05 m) from [tex]\( q_2 \)[/tex].
First, we need to determine the electric field contributions from both charges at the specified point.
1. Distance from the point to [tex]\( q_1 \)[/tex]:
Since the total distance between the charges is 0.25 m and the point is 0.05 m from [tex]\( q_2 \)[/tex],
[tex]\[ \text{Distance from point to } q_1 = 0.25 \, \text{m} - 0.05 \, \text{m} = 0.20 \, \text{m}. \][/tex]
2. Coulomb's Law:
The electric field [tex]\( E \)[/tex] created by a point charge [tex]\( q \)[/tex] at a distance [tex]\( r \)[/tex] is given by:
[tex]\[ E = \frac{k_e \cdot |q|}{r^2}, \][/tex]
where [tex]\( k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 \cdot \text{C}^{-2} \)[/tex].
3. Electric Field due to [tex]\( q_1 \)[/tex]:
[tex]\[ E_1 = \frac{(8.99 \times 10^9) \cdot (2.0 \times 10^{-7})}{(0.20)^2} = 44949.99999999999 \, \text{N/C}. \][/tex]
The direction of the electric field due to a positive charge is away from the charge.
4. Electric Field due to [tex]\( q_2 \)[/tex]:
[tex]\[ E_2 = \frac{(8.99 \times 10^9) \cdot (6.0 \times 10^{-8})}{(0.05)^2} = 215759.99999999994 \, \text{N/C}. \][/tex]
Since [tex]\( q_2 \)[/tex] is negative, the electric field direction is towards the charge.
5. Net Electric Field:
Both the electric fields [tex]\( E_1 \)[/tex] and [tex]\( E_2 \)[/tex] are along the same line between the charges. The total electric field at the point will be the vector sum of these fields. Since [tex]\( E_1 \)[/tex] is directed away from [tex]\( q_1 \)[/tex] and [tex]\( E_2 \)[/tex] is directed towards [tex]\( q_2 \)[/tex], they are in opposite directions.
Therefore, we can subtract the magnitudes to find the net electric field:
[tex]\[ E_{\text{net}} = E_1 - E_2 = 44949.99999999999 - 215759.99999999994 = -170809.99999999994 \, \text{N/C}. \][/tex]
The negative sign in the net electric field indicates that the direction of the net electric field is towards [tex]\( q_2 \)[/tex].
In summary:
- The electric field due to [tex]\( q_1 \)[/tex] is approximately [tex]\( 44950 \, \text{N/C} \)[/tex] away from [tex]\( q_1 \)[/tex].
- The electric field due to [tex]\( q_2 \)[/tex] is approximately [tex]\( 215760 \, \text{N/C} \)[/tex] towards [tex]\( q_2 \)[/tex].
- The net electric field at the point 5.0 cm from the negative charge is approximately [tex]\( -170810 \, \text{N/C} \)[/tex].
