Questions: Negative charge q1 is 1.00 m to the left of positive charge q2. In the figure below, determine the point (other than infinity) at which the electric field is zero. (Let q1=-3.30 μC and q2=7.00 μC.)

Negative charge q1 is 1.00 m to the left of positive charge q2. In the figure below, determine the point (other than infinity) at which the electric field is zero. (Let q1=-3.30 μC and q2=7.00 μC.)

Transcript text: Negative charge $q 1$ is 1.00 m to the left of positive charge $q 2$. In the figure below, determine the point (other than infinity) at which the electric field is zero. (Let $q 1=-3.30 \mu \mathrm{C}$ and $q^{2}=7.00 \mu \mathrm{C}$.)

Solution

Solution Steps

Step 1: Understand the Problem

We need to find the point along the line joining two charges where the electric field is zero. The charges are $q_1 = -3.30 \, \mu \mathrm{C}$ and $q_2 = 7.00 \, \mu \mathrm{C}$ , and they are 1.00 m apart. The electric field due to a point charge is given by $E = \frac{k|q|}{r^2}$ , where $k$ is Coulomb's constant, $q$ is the charge, and $r$ is the distance from the charge.

Step 2: Determine the Location of the Zero Electric Field

Since $q_1$ is negative and $q_2$ is positive, the point where the electric field is zero must be outside the segment joining the charges. It will be closer to the smaller magnitude charge, $q_1$ , because the electric field strength decreases with distance.

Step 3: Set Up the Equation for Zero Electric Field

Assume the point is at a distance $x$ to the left of $q_1$ . The distance from $q_2$ to this point is $x + 1.00 \, \mathrm{m}$ . The electric field due to $q_1$ at this point is $E_1 = \frac{k|q_1|}{x^2}$ and due to $q_2$ is $E_2 = \frac{k|q_2|}{(x + 1.00)^2}$ .