Two point charges are placed along a horizontal axis with the following values and positions: +3 µC at x = 0 cm and - 12 µC at x = 35 cm. At what point (the coordinate) along the x axis is the electric field zero? Expess the answer (in cm) with one decimal place.
43
a=43
To find the point along the x-axis where the electric field is zero, we can use the principle that the electric field due to a point charge decreases with distance and has the same sign as the charge itself.
Given that the charges are placed along a horizontal axis, the electric field due to the positive charge (+3 µC) will be directed away from it and the electric field due to the negative charge (-12 µC) will be directed towards it.
At some point between the charges, these electric fields will cancel out each other, resulting in a zero net electric field. Let's find that point.
Let's assume the point where the electric field is zero is at distance x from the positive charge (at x = 0 cm). Therefore, the distance from the negative charge (at x = 35 cm) would be (35 - x) cm.
Using Gauss's Law, we can calculate the electric field at these points, and set it equal to zero:
Electric field due to positive charge (E1): E1 = k * Q1 / r1^2
Electric field due to negative charge (E2): E2 = k * Q2 / r2^2
Where:
k is the electrostatic constant (k ≈ 9 x 10^9 Nm^2/C^2),
Q1 and Q2 are the charges (in Coulombs),
r1 is the distance between the point and the positive charge (in meters),
r2 is the distance between the point and the negative charge (in meters).
Since we are using centimeters for distances, we need to convert them to meters:
r1 = x / 100, and
r2 = (35 - x) / 100
Now, let's set E1 and E2 equal to each other and solve for x:
k * Q1 / r1^2 = k * Q2 / r2^2
Substituting the given values:
(9 x 10^9 Nm^2/C^2) * (3 x 10^-6 C) / (x/100)^2 = (9 x 10^9 Nm^2/C^2) * (-12 x 10^-6 C) / ((35 - x)/100)^2
Simplifying the equation:
(3 / x^2) = (-4 / (35 - x)^2)
Cross multiplying:
3 * (35 - x)^2 = -4 * x^2
Expanding and rearranging:
3 * (35^2 - 70x + x^2) = -4x^2
105^2 - 210x + 3x^2 = -4x^2
7x^2 - 210x + 1225 = 0
Dividing both sides by 7:
x^2 - 30x + 175 = 0
Using the quadratic formula:
x = (-(-30) ± √((-30)^2 - 4 * 1 * 175)) / (2 * 1)
Simplifying further:
x = (30 ± √(900 - 700)) / 2
x = (30 ± √200) / 2
x = 15 ± √50
Taking the positive value:
x = 15 + √50
Therefore, the point along the x-axis where the electric field is zero is at approximately 15 + 7.07 ≈ 22.1 cm.
To find the point along the x-axis where the electric field is zero, we can use the principle of superposition. The electric field at any point due to multiple charges is the vector sum of the electric fields created individually by each charge.
In this case, we have two charges (+3 µC and -12 µC) placed along the x-axis. Let's assume the point where the electric field is zero is at coordinate x = d cm.
The electric field at the point x = d cm due to the +3 µC charge can be calculated using Coulomb's law:
E1 = (k * q1) / (r1^2)
where k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2), q1 is the charge (+3 µC), and r1 is the distance between the +3 µC charge and the point x = d cm.
Similarly, the electric field at the same point due to the -12 µC charge is:
E2 = (k * q2) / (r2^2)
where q2 is the charge (-12 µC), and r2 is the distance between the -12 µC charge and the point x = d cm.
Since the electric field at the point x = d cm is zero, we can equate the two electric fields and solve for x:
E1 = E2
(k * q1) / (r1^2) = (k * q2) / (r2^2)
Now, let's substitute the values given in the problem:
(k * 3 µC) / (d^2) = (k * -12 µC) / ((35 - d)^2)
We can simplify this equation to solve for d.
12(d^2) = 3(35 - d)^2
Now, we can expand and solve for d.
12(d^2) = 3(1225 - 70d + d^2)
Simplifying further:
12d^2 = 3675 - 210d + 3d^2
Combining like terms:
9d^2 + 210d - 3675 = 0
Now, we can solve this quadratic equation for d using the quadratic formula:
d = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values:
d = (-(210) ± √((210)^2 - 4(9)(-3675))) / (2(9))
After solving this equation, we get two values for d: d1 and d2. The coordinates of the points where the electric field is zero are (d1, 0) and (d2, 0). The answer to the question is the value of d with one decimal place.
Note: It is recommended to use a calculator or software capable of solving quadratic equations to get the precise values for d1 and d2.