Question

Two particles with positive charges q1q1q_1 = 0.550 nCnC and q2q2q_2 = 8.50 nCnC are separated by a distance of 1.19 mm .

a.) At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?

Answer 1

Answer 2

To find the point along the line connecting the two charges where the total electric field is zero, we can use the principle of superposition. The electric field at any point is the vector sum of the electric fields produced by each individual charge.

Let's assume that the point along the line connecting the two charges, where the total electric field is zero, is at a distance x from charge q1 (with charge q1 = 0.550 nC) and at a distance (1.19 mm - x) from charge q2 (with charge q2 = 8.50 nC).

The electric field (E) due to a point charge can be calculated using Coulomb's law:

E = k * (|q| / r^2)

where: k is Coulomb's constant ≈ 8.99 x 10^9 N m^2/C^2, |q| is the magnitude of the charge, r is the distance from the charge to the point where we want to find the electric field.

At the point along the line where the total electric field is zero, the electric field due to q1 (E1) must be equal in magnitude but opposite in direction to the electric field due to q2 (E2). Mathematically:

E1 = E2

Now, let's set up the equations:

For charge q1: E1 = k * (|q1| / x^2)

For charge q2: E2 = k * (|q2| / (1.19 mm - x)^2)

Since E1 = E2, we can equate the two expressions:

k * (|q1| / x^2) = k * (|q2| / (1.19 mm - x)^2)

Now, plug in the known values:

0.550 nC = 8.50 nC / (1.19 mm - x)^2

Now, solve for x:

1.19 mm - x = √(8.50 nC / 0.550 nC) 1.19 mm - x = √(15.4545...) 1.19 mm - x ≈ 3.93 mm

Now, isolate x:

x ≈ 1.19 mm - 3.93 mm x ≈ -2.74 mm

The value of x is approximately -2.74 mm. Since distance cannot be negative, this result indicates that the point along the line where the total electric field is zero is approximately 2.74 mm on the other side of charge q1. In other words, it is approximately 1.19 mm + 2.74 mm ≈ 3.93 mm away from charge q2.