A point charge Q1=−2 μC is located at x=0, and a point charge Q2=+8 μC is placed at x=−0.5 m on the x-axis of a cartesian coordinate system.The goal of this problem is to determine the electric field, E⃗ (x)=E(x)xˆ, at various points along the x-axis.
(a)What is E(x) (in N/C) for x=-76.5 m ?
If you really want the help of an expert in this subject, you'll follow directions and put the SUBJECT in the School Subject box.
Ibn Zohr is a university in Morocco. which school you attend is of no interest on this forum. All we do is help students, so just the subject please. :)
To determine the electric field at a point, we can use the formula:
E(x) = k * (Q1 / r1^2) + k * (Q2 / r2^2)
where:
- E(x) is the electric field at point x
- k is the electrostatic constant (k ≈ 9 × 10^9 N m²/C²)
- Q1 and Q2 are the charges
- r1 and r2 are the distances between the charges
In this problem, we have Q1 = -2 μC, Q2 = +8 μC, and x = -76.5 m.
First, let's calculate the distances:
- r1 = -76.5 m (distance from x = -76.5 m to x = 0 m)
- r2 = -76.5 m - (-0.5 m) = -76 m (distance from x = -76.5 m to x = -0.5 m)
Plugging the values into the formula, we have:
E(x) = (9 × 10^9 N m²/C²) * (-2 μC) / (-76.5 m)^2 + (9 × 10^9 N m²/C²) * (+8 μC) / (-76 m)^2
Now, let's calculate the electric field:
E(x) = (9 × 10^9 N m²/C²) * (-2 × 10^-6 C) / (76.5 m)^2 + (9 × 10^9 N m²/C²) * (8 × 10^-6 C) / (76 m)^2
E(x) = -2.055 × 10^4 N/C + 2.705 × 10^4 N/C
E(x) = 0.65 × 10^4 N/C
Therefore, the electric field at x = -76.5 m is 0.65 × 10^4 N/C.
To find the electric field at a given point, we can use the formula:
E(x) = k * (Q / r^2)
Where:
- E(x) is the electric field at the point with coordinate x.
- k is the electrostatic constant, approximately equal to 9 * 10^9 N m^2/C^2.
- Q is the charge that produces the electric field.
- r is the distance between the point and the charge.
In this case, there are two charges present on the x-axis:
- Q1 = -2 μC located at x = 0.
- Q2 = +8 μC located at x = -0.5 m.
Since the electric field is a vector quantity, we need to consider the direction as well. Here, the electric field will be directed towards the charge Q1 at x = 0.
Now, let's calculate the electric field at x = -76.5 m.
For this point, we need to find the electric field due to both charges and then add them up since the electric field is a vector sum.
First, let's calculate the electric field due to Q1 at x = -76.5 m:
r1 = x - x1 = -76.5 m - 0 m = -76.5 m
E1 = k * (Q1 / r1^2)
= (9 * 10^9 N m^2/C^2) * (-2 * 10^-6 C) / (-76.5 m)^2
Next, let's calculate the electric field due to Q2 at x = -76.5 m:
r2 = x - x2 = -76.5 m - (-0.5 m) = -76 m
E2 = k * (Q2 / r2^2)
= (9 * 10^9 N m^2/C^2) * (8 * 10^-6 C) / (-76 m)^2
Finally, we can calculate the total electric field by summing the contributions from both charges:
E(x) = E1 + E2
With the calculated values of E1 and E2, you can simply add them up to get the total electric field at x = -76.5 m in N/C.