Question

Two particles with charges of +5.70 nC and −8.55 nC are separated by 2.40 m. (a) What is the magnitude of the electrostatic force between the particles? N

Answer 1

Answer:

The magnitude of the electrostatic force is $7.615\times10^{-8}\ N$

Explanation:

Given that,

Charge $q_{1}= 5.70\ nC$

Charge $q_{2}=-8.55\ nC$

Distance = 2.40 m

We need to calculate the electrostatic force

Using formula of electrostatic force

$F= \dfrac{kq_{1}q_{2}}{d^2}$

Where, k = coulomb's constant

q = magnitude of the charge

d = distance

Put the value into the formula

$F=\dfrac{9\times10^{9}\times5.70\times10^{-9}\times8.55\times10^{-9}}{(2.40)^2}$

$F=7.615\times10^{-8}\ N$

Hence, The magnitude of the electrostatic force is $7.615\times10^{-8}\ N$

Answer 2

The electrostatic force between the two particles is 76.14nN

Data;

• q1 = 5.70 nC
• q2 = -8.55 nC
• r = 2.40

Electrostatic Force

To find the electrostatic force between the two charges, we would apply coulomb's law which is given as

$F = \frac{kq_1q_2}{r^2}$

Let's substitute the values and solve.

$F = \frac{9*10^9 * 5.70*10^-^9 * 8.55*10^-^9}{2.4^2}\\ F = 7.61*10^-8N\\F = 76.14nN$

The electrostatic force between the two particles is 76.14nN

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