Two charges are located along the x-axis, qA = -4 C at point A and qB = 7 C at point B. They are separated by a distance of d = 4 m. There is some point P along the x-axis at which the net Electric potential is zero.

What is the distance from point A to point P? 
 m

What is the distance from point B to point P? 
 m

To find the distance from points \( A \) and \( B \) to the point \( P \) where the net electric potential is zero, we need to consider the principle that the total electric potential at a point due to multiple charges is the algebraic sum of the individual potentials.

### Formula for Electric Potential:
The electric potential \( V \) at a point due to a charge \( q \) at a distance \( r \) is given by:
\[
V = \frac{kq}{r}
\]
where:
- \( V \) is the electric potential,
- \( k \) is Coulomb's constant (\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)),
- \( q \) is the charge,
- \( r \) is the distance from the charge.

Now, let's assume point \( P \) is a distance \( x \) from charge \( q_A \) (at point A). Since both charges are along the x-axis and are separated by \( d = 4 \, \text{m} \), the distance from \( q_B \) to point \( P \) would be \( (4 - x) \).

### Condition for Zero Potential:
For the net potential at point \( P \) to be zero, the potentials due to the charges at point A and point B must add up to zero. Thus:
\[
\frac{kq_A}{x} + \frac{kq_B}{4 - x} = 0
\]

### Substituting the Charges:
The charges are:
- \( q_A = -4 \, \text{C} \),
- \( q_B = 7 \, \text{C} \).

Thus, the equation becomes:
\[
\frac{-4k}{x} + \frac{7k}{4 - x} = 0
\]
The Coulomb constant \( k \) cancels out, simplifying the equation to:
\[
\frac{-4}{x} + \frac{7}{4 - x} = 0
\]

### Solving the Equation:
1. Multiply through by \( x(4 - x) \) to eliminate the denominators:
\[
-4(4 - x) + 7x = 0
\]
2. Expand and simplify:
\[
-16 + 4x + 7x = 0
\]
\[
11x = 16
\]
\[
x = \frac{16}{11} \approx 1.45 \, \text{m}
\]

So, the distance from point A to point P is approximately \( 1.45 \, \text{m} \).

### Distance from Point B to Point P:
Since the total distance between \( A \) and \( B \) is \( 4 \, \text{m} \), the distance from point \( B \) to point \( P \) is:
\[
4 - 1.45 = 2.55 \, \text{m}
\]

### Final Answers:
- The distance from point \( A \) to point \( P \) is \( 1.45 \, \text{m} \).
- The distance from point \( B \) to point \( P \) is \( 2.55 \, \text{m} \).

Would you like more details on any step or have any questions?

---

Here are some related questions:
1. What happens to the electric potential at point P if the charges are swapped?
2. How does changing the distance between the charges affect the location of point P?
3. What if both charges had the same sign? Where would the zero potential point be?
4. How would you calculate the electric field at point P?
5. What if point P was not along the x-axis? How would you approach the problem?

**Tip**: When solving problems involving electric potential, always consider the signs of the charges, as they dictate the direction of the potential contribution.

Two charges are located along the x-axis, qA = -4 C at point A and qB = 7 C at point B. They are separated by a distance of d = 4 m. There is some point P along the x-axis at which the net Electric potential is zero.

What is the distance from point A to point P?
 m

What is the distance from point B to point P?
 m

Learn how to find the point where the net electric potential is zero between two charges on the x-axis. This problem involves calculating the electric potential from two charges qA = -4 C and qB = 7 C, separated by 4 meters, and determining the distances from point A and point B to point P. Ideal for students in advanced physics classes.

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