Post

Created by @nathanedwards
 at November 4th 2023, 10:28:26 pm.

Electric Force and Field Question:

Two point charges, +Q and -2Q, are placed 3 meters apart in a vacuum. Calculate the electric force between them.

Answer:

To calculate the electric force between two point charges, we can use Coulomb's Law formula:

F=kq1q2r2 F = k \frac{{|q_1 q_2|}}{{r^2}}

where:

  • F is the electric force between the charges,
  • k is the electrostatic constant (k=8.99×109Nm2/C2k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2),
  • q1q_1 and q2q_2 are the magnitudes of the two charges, and
  • r is the separation distance between the charges.

In this case, we have a positive charge, QQ, and a negative charge, 2Q-2Q, placed 3 meters apart. Let's calculate the electric force using Coulomb's Law:

F=kQ(2Q)(3m)2 F = k \frac{{|Q(-2Q)|}}{{(3 \, \text{m})^2}}

We can simplify this expression:

F=8.99×109Nm2/C2×2Q29m2 F = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \times \frac{{2Q^2}}{{9 \, \text{m}^2}}

Rearranging and canceling units, we get:

F=2×(8.99×109)×Q29N F = \frac{{2 \times (8.99 \times 10^9) \times Q^2}}{{9}} \, \text{N}

Therefore, the electric force between the two charges is:

F=(1.998×1010)×Q29N F = \frac{{(1.998 \times 10^{10}) \times Q^2}}{{9}} \, \text{N}

Note: This is the magnitude of the force. The direction of the force would depend on the sign of the charges.