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Sagot :
To find the point on the x-axis where the net electric field due to two charges is zero, we need to consider the electric fields produced by each charge and set their net sum to zero at that particular point.
Here's a step-by-step breakdown of the problem:
1. Understanding the setup:
- Charge [tex]\( q_1 = 0.96 \, \mu C \)[/tex] is located at [tex]\((d, 0)\)[/tex].
- Charge [tex]\( q_2 = 2.47 \, \mu C \)[/tex] is located at [tex]\((-d, 0)\)[/tex].
- The distance [tex]\( d \)[/tex] is [tex]\( 0.383 \, m \)[/tex].
2. Electric field due to a point charge:
The electric field [tex]\( E \)[/tex] due to a charge [tex]\( q \)[/tex] at a distance [tex]\( r \)[/tex] is given by:
[tex]\[ E = \frac{k \cdot q}{r^2} \][/tex]
where [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)[/tex]).
3. Setting up the equations:
Considering a point [tex]\( x_0 \)[/tex] on the x-axis:
- The electric field at [tex]\( (x_0, 0) \)[/tex] due to [tex]\( q_1 \)[/tex] located at [tex]\( (d, 0) \)[/tex]:
[tex]\[ E_1 = \frac{k \cdot q_1}{(x_0 - d)^2} \][/tex]
- The electric field at [tex]\( (x_0, 0) \)[/tex] due to [tex]\( q_2 \)[/tex] located at [tex]\( (-d, 0) \)[/tex]:
[tex]\[ E_2 = \frac{k \cdot q_2}{(x_0 + d)^2} \][/tex]
4. Net electric field is zero:
For the net electric field to be zero at [tex]\( (x_0, 0) \)[/tex], the magnitudes of the electric fields due to both charges should be equal:
[tex]\[ E_1 = E_2 \][/tex]
5. Setting the magnitudes equal and canceling out constants:
[tex]\[ \frac{k \cdot q_1}{(x_0 - d)^2} = \frac{k \cdot q_2}{(x_0 + d)^2} \][/tex]
[tex]\[ \frac{q_1}{(x_0 - d)^2} = \frac{q_2}{(x_0 + d)^2} \][/tex]
6. Substituting the values:
- [tex]\( q_1 = 0.96 \times 10^{-6} \, C \)[/tex]
- [tex]\( q_2 = 2.47 \times 10^{-6} \, C \)[/tex]
- [tex]\( d = 0.383 \, m \)[/tex]
The equation becomes:
[tex]\[ \frac{0.96 \times 10^{-6}}{(x_0 - 0.383)^2} = \frac{2.47 \times 10^{-6}}{(x_0 + 0.383)^2} \][/tex]
7. Solving the equation:
Rewriting the equation, we get:
[tex]\[ \frac{0.96}{(x_0 - 0.383)^2} = \frac{2.47}{(x_0 + 0.383)^2} \][/tex]
This simplifies and solves to find [tex]\( x_0 \)[/tex].
8. Solution:
Solving the above equation, we find that:
[tex]\[ x_0 = 1.65114603661292 \, \text{meters} \][/tex]
Therefore, the value of [tex]\( x_0 \)[/tex] in meters is:
[tex]\[ x_0 = 1.651 \, \text{meters} \][/tex]
Please submit this final answer in the provided box.
Here's a step-by-step breakdown of the problem:
1. Understanding the setup:
- Charge [tex]\( q_1 = 0.96 \, \mu C \)[/tex] is located at [tex]\((d, 0)\)[/tex].
- Charge [tex]\( q_2 = 2.47 \, \mu C \)[/tex] is located at [tex]\((-d, 0)\)[/tex].
- The distance [tex]\( d \)[/tex] is [tex]\( 0.383 \, m \)[/tex].
2. Electric field due to a point charge:
The electric field [tex]\( E \)[/tex] due to a charge [tex]\( q \)[/tex] at a distance [tex]\( r \)[/tex] is given by:
[tex]\[ E = \frac{k \cdot q}{r^2} \][/tex]
where [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)[/tex]).
3. Setting up the equations:
Considering a point [tex]\( x_0 \)[/tex] on the x-axis:
- The electric field at [tex]\( (x_0, 0) \)[/tex] due to [tex]\( q_1 \)[/tex] located at [tex]\( (d, 0) \)[/tex]:
[tex]\[ E_1 = \frac{k \cdot q_1}{(x_0 - d)^2} \][/tex]
- The electric field at [tex]\( (x_0, 0) \)[/tex] due to [tex]\( q_2 \)[/tex] located at [tex]\( (-d, 0) \)[/tex]:
[tex]\[ E_2 = \frac{k \cdot q_2}{(x_0 + d)^2} \][/tex]
4. Net electric field is zero:
For the net electric field to be zero at [tex]\( (x_0, 0) \)[/tex], the magnitudes of the electric fields due to both charges should be equal:
[tex]\[ E_1 = E_2 \][/tex]
5. Setting the magnitudes equal and canceling out constants:
[tex]\[ \frac{k \cdot q_1}{(x_0 - d)^2} = \frac{k \cdot q_2}{(x_0 + d)^2} \][/tex]
[tex]\[ \frac{q_1}{(x_0 - d)^2} = \frac{q_2}{(x_0 + d)^2} \][/tex]
6. Substituting the values:
- [tex]\( q_1 = 0.96 \times 10^{-6} \, C \)[/tex]
- [tex]\( q_2 = 2.47 \times 10^{-6} \, C \)[/tex]
- [tex]\( d = 0.383 \, m \)[/tex]
The equation becomes:
[tex]\[ \frac{0.96 \times 10^{-6}}{(x_0 - 0.383)^2} = \frac{2.47 \times 10^{-6}}{(x_0 + 0.383)^2} \][/tex]
7. Solving the equation:
Rewriting the equation, we get:
[tex]\[ \frac{0.96}{(x_0 - 0.383)^2} = \frac{2.47}{(x_0 + 0.383)^2} \][/tex]
This simplifies and solves to find [tex]\( x_0 \)[/tex].
8. Solution:
Solving the above equation, we find that:
[tex]\[ x_0 = 1.65114603661292 \, \text{meters} \][/tex]
Therefore, the value of [tex]\( x_0 \)[/tex] in meters is:
[tex]\[ x_0 = 1.651 \, \text{meters} \][/tex]
Please submit this final answer in the provided box.
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