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3.What are the forces (in N) on two charges of +0.800 C and −5.00 C, respectively, if they are separated by a distance of 5.00 m?magnitude answer in:_____Nwill it: attract or repel?

3 What are the forces in N on two charges of 0 800 C and 5 00 C respectively if they are separated by a

Answer :

Final answer:

The force between two charges +0.800 C and -5.00 C separated by 5.00 m is calculated using Coulomb's Law. The charges will attract each other because they are of opposite signs. The formula considers the absolute value of their product.

Explanation:

The forces on two charges can be calculated using Coulomb's Law, which states that the force (F) between two charges (q₁ and q₂) is proportional to the product of the charges and inversely proportional to the square of the distance (r) between them. The formula is given by:

F = k * |q1 * q2| / r²

where k is Coulomb's constant (8.99 x 10¹ Nm²/C²).

For charges +0.800 C and −5.00 C separated by 5.00 m, considering the sign of the charges, they will attract each other because one is positive and the other is negative. By plugging values into Coulomb's Law:

F = (8.99 x 10¹) * (0.800 C * 5.00 C) / (5.00 m)²

Note: Since the charges have opposite signs, the equation considers the absolute value of their product.

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Rewritten by : Barada

Given:

The charge q1 = 0.8 C

The charge q2 = -5 C

The distance between the charges is r = 5 m

Required: The magnitude of force and type of force (attraction/repulsion)

Explanation:

The magnitude of the force can be calculated by Coulomb's law

[tex]F=k\frac{|q1q2|}{r^2}[/tex]

Here, k is the electrostatic constant whose value is

[tex]k=9\times10^9\text{ Nm}^2\text{ /C}^2[/tex]

On substituting the values, the magnitude of force will be

[tex]\begin{gathered} F=9\times10^9\times\frac{0.8\times5}{(5)^2} \\ =1.44\times10^9\text{ N} \end{gathered}[/tex]

The charges are opposite in nature, thus the charges will attract.

Final Answer: The magnitude of the force is 1.44e9 N and the charges will attract.