Question

A charge of 4.5 × 10-9 C is located 3.2 m from a charge of -2.8 × 10-9 C. Find the

electrostatic force exerted by one charge on the other.

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Answer 1

Answer:

$F = -1.107*10^{-8} N\\|F| = 1.107*10^{-8} N$

Explanation:

$q_1 = 4.5 * 10^{-9} C$

$q_2 = -2.8 * 10^{-9} C$

The distance separating the two charges, r = 3.2 m

According to Coulomb's law of electrostatic attraction, the electrostatic force between the two charges can be given by the formula:

$F = \frac{kq_{1} q_{2} }{r^2}$

Where $k = 9.0 * 10^9 Nm^2/C^2$

$F = \frac{9*10^9 * 4.5*10^{-9} * (-2.8*10^{-9}}{3.2^2} \\F = \frac{-113.4*10^{-9}}{10.24}\\F = -11.07 *10^{-9}\\F = -1.107*10^{-8}N$

Answer 2

Answer:

F = 1.1074 × $10^{-8}$ N

Explanation:

An electrostatic force is either a force or attraction or repulsion between two charges. It can be determined by:

F = $\frac{kq_{1}q_{2} }{r^{2} }$

where: F is the force, k is a constant, $q_{1}$ is the first charge, $q_{2}$ is the second charge and r the distance between the charges.

Given that: k = 9 × $10^{9}$ N$m^{2}$$C^{-2}$, $q_{1}$ = 4.5 × $10^{-9}$C, $q_{2}$ = -2.8 × $10^{-9}$C and r = 3.2 m.

Then,

F = $\frac{9*10^{9}*4.5*10^{-9}*2.8*10^{-9} }{3.2^{2} }$

 = $\frac{1.134*10^{-7} }{10.24}$

 = 1.1074 × $10^{-8}$

The electrostatic force exerted is 1.1074 × $10^{-8}$ N, and it is a force of attraction.