Question

Two objects have charges 2.0 C and 1.0 C. If the objects are placed 2 meters apart, what is the magnitude of the force that the 2 C object exerts on the 1 C obejct?

Answer 1

6. The magnitude of the force that the 2 C object exerts on the 1 C object is 4.495 * 10^9 N. Option C is the right choice.

7. Only increasing the combined charge of both spheres increases the electrical force between them. Option C is the right choice.

6.

The force between two charged objects is given by Coulomb's law:

F = k * |q1 * q2| / r^2

where:

F is the force between the two charges

k is Coulomb's constant, approximately equal to 8.99 * 10^9 N⋅m^2/C^2

q1 and q2 are the magnitudes of the charges of the two objects

r is the distance between the two objects

In this case, q1 = 2.0 C, q2 = 1.0 C, and r = 2 meters. Plugging these values into Coulomb's law, we get:

F = 8.99 * 10^9 N⋅m^2/C^2 * |2.0 C * 1.0 C| / (2 meters)^2

F = 4.495 * 10^9 N

So the answer is Option C. 4.495 * 10^9N.

7.

The electrical force between two charged spheres depends on the magnitude of their charges and the distance between them, as described by Coulomb's Law.

Increasing the combined mass of the spheres, the diameter of one sphere, or the distance between them has no effect on the force.

Therefore, the only option that increases the force is to increase the combined charge of both spheres. Option C is the right choice.

Answer 2

`4.5\cdot 10^9 N`

Step-by-step explanation:

The electric force between two charged objects is given by:

`F=k(q_1 q_2)/(r^2)`

where:

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is their separation

In this problem:

q1 = 2.0 C

q2 = 1.0 C

r = 2 m

So, the electric force is

`F=(9\cdot 10^9 Nm^2C^(-2))((2.0 C)(1.0 C))/((2 m)^2)=4.5\cdot 10^9 N`