Two point charges are 2.00 m apart as shown on the right. The charge of the first object is -1.50 C, while the second object has a charge of +6.85 C. At which point will the electric field vector be zero?

Question

in simple steps

Answer 1

To find the point where the electric field vector is zero, we can use the principle of superposition. According to this principle, the electric field at a point due to multiple charges is the vector sum of the electric fields at that point due to each individual charge.

1. First, let's calculate the electric field at the point P1 (to the right of the -1.50 C charge). The electric field due to the -1.50 C charge can be calculated using Coulomb's law:
E1 = k * q1 / r1^2
where k is the Coulomb's constant (approximately 9 * 10^9 N m^2/C^2), q1 is the charge of the -1.50 C object, and r1 is the distance between the -1.50 C object and the point P1.

2. Next, let's calculate the electric field at the point P2 (to the left of the +6.85 C charge). The electric field due to the +6.85 C charge can be calculated using Coulomb's law:
E2 = k * q2 / r2^2
where q2 is the charge of the +6.85 C object, and r2 is the distance between the +6.85 C object and the point P2.

3. Now, we need to find the point where the electric field vectors cancel each other out. This means that the vector sum of E1 and E2 should be zero:
E1 + E2 = 0

4. Rearranging the equation, we have:
E1 = -E2

5. Substituting the expressions for E1 and E2 from step 1 and step 2 respectively:
k * q1 / r1^2 = -k * q2 / r2^2

6. Simplifying, we have:
q1 / r1^2 = -q2 / r2^2

7. Rearranging the equation, we have:
q1 / q2 = -r1^2 / r2^2

8. Substituting the given values: q1 = -1.50 C, q2 = +6.85 C, r1 = 2.00 m, and r2 = 2.00 m:
-1.50 C / 6.85 C = -4.00 m^2 / 2.00 m^2

9. Simplifying, we have:
-0.2197 ≈ -0.25 ≈ -1/4

10. From step 9, we can conclude that the ratios of distances between the -1.50 C charge and the point P1, and between the +6.85 C charge and the point P2 are in the ratio of 1:2. Therefore, the point where the electric field vector is zero is 1/3 of the way from the -1.50 C charge towards the +6.85 C charge.

So, the electric field vector will be zero at a point 2/3 of the distance from the -1.50 C charge towards the +6.85 C charge.