Question

A negative charge of - 4.0 x 10-5 C and a positive charge of

7.0 x 10 -5 C are separated by 0.15 m. What is the force

between the two charges?

Answer 1

Answer:

1117.51N/C

Explanation:

The magnitude of the electric force is given by:

$|\vec{F}|=|k\frac{q_1q_2}{r^2}|$

k: Coulomb's constant = 8.98*10^9Nm^/2C^2

r: distance between the charges = 0.15m

By replacing the values of q1, q2, k and r you obtain:

$|\vec{F}|=|(8.98*10^9Nm^2/C^2)\frac{(-4.0*10^{-5}C)(7.0*10^{-5}C)}{(0.15m)^2}|=1117.51\frac{N}{C}$

hence, the force between the charges is 1117.51N/C

Answer 2

Answer:

The force of attraction between the charges is 1120 N

Explanation:

Given:

positive charge of the particle, q1 = 7 x 10^-5 C

negative charge of the particle, q2 = -4 x 10^-5 C

distance between the charges, r = 0.15 m

The force of attraction between the charges will be calculated using Coulomb's law:

F = (k|q1q2|) / r^2

Where:

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 Nm^2/c^2

|q1| is magnitude of charge 1

|q2| is magnitude of charge 2

F = (9 x 10^9 x 7 x 10^-5 x 4 x 10^-5) / (0.15 x 0.15)

F = 1120 N

Thus, the force between the charges is 1120 N