Question

Two identical conducting spheres carry charges of +5.0 μC and –1.0 μC, respectively. The centers of the spheres are initially separated by a distance L. The two spheres are brought together so that they are in contact. The spheres are then returned to their original separation L. What is the ratio of the magnitude of the electric force on either sphere after the spheres are touched to that before they were touched? A) 1/1 B) 4/5 C) 9/5 D) 5/1 E) 4/9

Answer 1

After two identical conducting spheres carrying different charges come into contact and separate, their total charge is distributed evenly. Using Coulomb's Law, the ratio of the forces after and before contact is 4/5. The answer is B) 4/5.

Step-by-step explanation:

When two identical conducting spheres come into contact, their total charge is redistributed evenly between them. In this case, we have one sphere with a charge of +5.0 μC and another with –1.0 μC, totaling +4.0 μC for both spheres. After contact and redistribution, each sphere will have half of the total charge, which is +2.0 μC per sphere.

To find the ratio of the magnitudes of the electric forces before and after contact, we use Coulomb's Law, which states that the electric force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between their centers:F = k * |q1 * q2| / L2

Where F is the force, k is Coulomb's constant, q1 and q2 are the charges, and L is the separation distance. Before contact, the force is: Fbefore = k * |(+5.0 μC) * (–1.0 μC)| / L2 = k * 5.0 μC / L2

After contact, the force is: Fafter = k * |(+2.0 μC) * (+2.0 μC)| / L2 = k * 4.0 μC / L2

The ratio of Fafter to Fbefore is therefore (4.0/5.0 μC), which simplifies to 4/5. The answer to the question is option B) 4/5.

Answer 2

B) 4/5

Step-by-step explanation:

The magnitude of the electric force between the two spheres is given by

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

where

k is the Coulombs' constant

q1 and q2 are the charges on the two spheres

r is the distance between the two spheres

Initially, we have

`q_1 = 5.0\mu C=5.0\cdot 10^(-6)C\\q_2 = 1.0 \mu C=1.0 \cdot 10^(-6)C\\r = L`

So the force is

`F_1=k((5.0\cdot 10^(-6)C)(1.0\cdot 10^(-6)C))/(L^2)=(5.0 \cdot 10^(-12) C^2)(k)/(L^2)`

Later, the two spheres are brought together so they are in contact: this means that the total charge will redistribute equally on the two spheres (because they are identical).

The total charge is

`Q=q_1 + q_2 = +4.0 \mu C=4.0\cdot 10^(-6)C`

So each sphere will have a charge of

`q=(Q)/(2)=2.0\cdot 10^(-6) C`

So, the new force will be

`F_2=k((2.0\cdot 10^(-6)C)(120\cdot 10^(-6)C))/(L^2)=(4.0 \cdot 10^(-12) C^2)(k)/(L^2)`

And so the ratio of the two forces is

`(F_2)/(F_1)=(4.0\cdot 10^(-12) C)/(5.0\cdot 10^(-12) C)=(4)/(5)`