Two charges, -2.0 micro C and -6.0 micro C, are located at (-0.65 m, 0) and (0.65 m, 0), respectively. There is a point on the x-axis between the two charges where the electric field is zero. Find the location of the point where the electric field is zero.
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To find the location where the electric field is zero, we need to use the principle of superposition. According to this principle, the total electric field at a point due to multiple charges is the vector sum of the electric fields produced by each individual charge.
Given that we have two charges, -2.0 μC and -6.0 μC, we can calculate the electric fields produced by each charge separately at a point on the x-axis.
Let's denote the distance of the point from the first charge (-2.0 μC) as x1 and the distance from the second charge (-6.0 μC) as x2.
The electric field produced by a point charge is given by the equation:
E = k * (q / r^2)
where E is the electric field, k is the electrostatic constant (approximately 9.0 x 10^9 N·m^2/C^2), q is the charge, and r is the distance between the charge and the point.
For the first charge (-2.0 μC), the electric field at the point is:
E1 = k * (q1 / r1^2)
Using the values given, q1 = -2.0 μC, r1 = x1, and substituting the values into the equation, we have:
E1 = k * (-2.0 μC) / x1^2
Similarly, for the second charge (-6.0 μC), the electric field at the point is:
E2 = k * (q2 / r2^2)
Using the values given, q2 = -6.0 μC, r2 = x2, and substituting the values into the equation, we have:
E2 = k * (-6.0 μC) / x2^2
Now, since the electric field is a vector quantity, it has both magnitude and direction. The direction of the electric field at the point between the charges will be opposite for each charge since they are both negative charges. As a result, we need to consider the signs when calculating the total electric field.
To find the point where the electric field is zero, the magnitudes of E1 and E2 should be equal:
|E1| = |E2|
Mathematically, we can express this as:
k * (2.0 μC) / x1^2 = k * (6.0 μC) / x2^2
Canceling out the constants and rearranging the equation, we get:
(x2 / x1)^2 = 3
Taking the square root of both sides, we have:
x2 / x1 = √3
Now, we know that x1 + x2 = 1.3 m (since the sum of the distances from the charges is the total length of 1.3 m).
By substituting the value of x2 / x1 as √3 into this equation, we can solve for x1:
x1 + √3x1 = 1.3
Factoring out x1, we get:
x1(1 + √3) = 1.3
Dividing both sides by (1 + √3), we get:
x1 = 1.3 / (1 + √3)
Evaluating this expression, we find:
x1 ≈ 0.358 m
Therefore, the location on the x-axis where the electric field is zero is approximately 0.358 m from the charge of -2.0 μC.