A positive point charge = 2.1×10−5 is fixed at the origin of coordinates, and a negative point charge = −5.4×10−6 is fixed to the axis at .
abeer abdullah
Part A
Find the location of the place along the axis where the electric field due to these two charges is zero.
Express your answer using two significant
Right of charge
To find the location along the x-axis where the electric field due to these two charges is zero, we need to consider the electric field equations for point charges.
The electric field produced by a point charge is given by:
E = k * q / r^2
Where
E is the electric field,
k is the Coulomb's constant (approximated as 8.99 x 10^9 Nm^2/C^2),
q is the charge, and
r is the distance from the charge.
In this case, we have a positive charge of +2.1 x 10^-5 C at the origin (0,0) and a negative charge of -5.4 x 10^-6 C located on the x-axis (x,0).
Let's assume the distance of the zero electric field point from the origin is represented by "x".
The electric field due to the positive charge at the zero electric field point is given by:
E1 = (8.99 x 10^9 Nm^2/C^2) * (2.1 x 10^-5 C) / (x)^2
The electric field due to the negative charge at the zero electric field point is given by:
E2 = (8.99 x 10^9 Nm^2/C^2) * (-5.4 x 10^-6 C) / (x - a)^2
(Where "a" represents the position of the negative charge on the x-axis, which is given as 0.25 m)
For the electric field to be zero at a particular point, the magnitudes of the electric fields produced by the positive and negative charges must be equal:
E1 = E2
Substituting the values of E1 and E2 and solving for x:
(8.99 x 10^9 Nm^2/C^2) * (2.1 x 10^-5 C) / (x)^2 = (8.99 x 10^9 Nm^2/C^2) * (-5.4 x 10^-6 C) / (x - 0.25)^2
Simplifying this equation, we get:
(2.1 x 10^-5) / (x)^2 = (-5.4 x 10^-6) / (x - 0.25)^2
Now, we can multiply both sides of the equation by (x)^2 to eliminate the denominators:
2.1 x 10^-5 = (-5.4 x 10^-6) * (x)^2 / (x - 0.25)^2
Cross-multiplying and simplifying further, we get:
2.1 x 10^-5 * (x - 0.25)^2 = (-5.4 x 10^-6) * (x)^2
Expanding both sides of the equation:
2.1 x 10^-5 * (x^2 - 0.5x + 0.0625) = -5.4 x 10^-6 * (x^2)
Multiplying out and rearranging, we get a quadratic equation:
2.1 x 10^-5 * x^2 - 1.05 x 10^-5 * x + 1.3125 x 10^-6 = -5.4 x 10^-6 * x^2
2.1 x 10^-5 * x^2 - 5.4 x 10^-6 * x^2 - 1.05 x 10^-5 * x + 1.3125 x 10^-6 = 0
Combining the like terms:
(2.1 - 5.4) x 10^-5 * x^2 - 1.05 x 10^-5 * x + 1.3125 x 10^-6 = 0
Simplifying further:
-3.3 x 10^-5 * x^2 - 1.05 x 10^-5 * x + 1.3125 x 10^-6 = 0
This is a quadratic equation. We can solve it using the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / (2a)
In the equation -3.3 x 10^-5 * x^2 - 1.05 x 10^-5 * x + 1.3125 x 10^-6 = 0, we can identify:
a = -3.3 x 10^-5,
b = -1.05 x 10^-5, and
c = 1.3125 x 10^-6.
Substituting these values into the formula and solving for x will give us the location along the x-axis where the electric field due to these two charges is zero.