A 5.0-ìC point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-ìC point charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field due to these charges equal to zero?

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To find the point on the line where the electric field due to the charges is zero, we can use the principle of superposition. The electric field at any point on the line is the vector sum of the electric fields due to each charge individually.

Let's assume that the point on the line between the charges is at a distance x from the 5.0 μC point charge (located at the 0.00 cm mark).

The electric field due to the 5.0 μC charge is given by Coulomb's law:

E1 = k * q1 / r1^2

where k is the Coulomb's constant (9 × 10^9 Nm^2/C^2), q1 is the charge (5.0 μC), and r1 is the distance from the charge to the point on the line (x).

Similarly, the electric field due to the -4.0 μC charge (located at the 50 cm mark) is:

E2 = k * q2 / r2^2

where q2 is the charge (-4.0 μC), and r2 is the distance from the charge to the point on the line (50 cm - x).

In order for the total electric field to be zero at the point on the line, E1 + E2 = 0.

Substituting the values and rearranging the equation, we get:

k * q1 / r1^2 + k * q2 / r2^2 = 0

Plug in the given values:

(9 × 10^9 Nm^2/C^2) * (5.0 × 10^-6 C) / x^2 + (9 × 10^9 Nm^2/C^2) * (-4.0 × 10^-6 C) / (50 cm - x)^2 = 0

Now, you can solve this equation for x. Rearrange the equation, combine the terms, and then solve for x using algebraic methods.

Note: Make sure to convert the distance from cm to meters if necessary for consistent units in the equation.

Once you solve for x, you will find the point on the line where the electric field due to these charges is zero.