Question

Two thin hollow plastic spheres, about the size of a ping-pong ball with masses (), have been rubbed with wool. Sphere 1 has a charge ofand is at location. Sphere 2 has a charge ofand is atlocation. It will be useful to draw a diagram of the situation, including the relevant vectors.

a) What is the relative position vectorpointing fromto? b) What is the distance betweenand? c) What is the unit vectorin the direction of? d) What is the magnitude of the gravitational force exerted onby? e) What is the (vector) gravitational force exerted onby? f) What is the magnitude of the electric force exerted onby? g) What is the (vector) electric force exerted onby? h) What is the ratio of the magnitude of the electric force to the magnitude of the gravitational force? i) if the two masses were four times further away (that is, if the distance between the masses were), what would be the ratio of the magnitude of the electric force to the magnitude of the gravitational force now?

Short answer

• a) The relative position vector is.

• b) the distance between the pointsis.

• c) the unit vector is.

• d) the magnitude of the gravitational force is.

• e) the (vector) gravitational force exerted onis.

• f) the magnitude of the electric force exerted onis.

• g) the (vector) electric force exerted onis.

• h) the ratio of the magnitude of the electric force to the magnitude of the gravitational force is.

• i) The ratio of the magnitude of the electric force and the gravitational force is.

Step-by-step solution

Identification of the given data

• The mass of the hollow plastic spheres is.

• The charge of sphere 1 is.

• The first sphere is at the location.

• The charge of sphere 2 is.

• The first sphere is at the location.

Significance of the Newton’s gravitational NS Coulomb’s law in identifying the forces

Newton’s gravitational law states that the particle in the field of another particle feels a force that is directly proportional to the product of the mass and inversely proportional to the square of the distance.

Coulomb’s law states that the unlike charges attract and like charges repel each other.

The equation of the gravitational and Coulomb’s equation give the gravitational and the electric force along with the position, distance, unit vector, and the ratio of the gravitational and electric force.

Determination of the gravitational and the electric force along with the position and the unit vector and the ratio of the electric and gravitational forces

The free-body diagram of the two particles and the vectors have been provided below.

a) According to Newton’s laws, the position vector for the sphere 1 can be expressed as:

Similarly, the position vector for the sphere 2 can be expressed as:

Hence, the position vector fromis described as:

Thus, the relative position vector is.

b) The position vector can help calculate the distance between the particles.

Hence, the distance between the pointscan be expressed as:

Substituting the values in the above equation, we get-

Thus, the distance between the pointsis.

c) From Newton’s gravitational law, the unit vector can be expressed as:

Here,is the unit vector,is the position vector, andis the distance between the points.

Substituting the values in the above equation and using the value from the equation, we get-

Thus, the unit vector is.

d) From Newton’s gravitational law, the gravitational force exerted on the spherecan be expressed as:

Here, F is the gravitational force, G is the gravitational constant,are the mass of the spheres that are,andis the distance amongst them.

Substituting the values in the above equation, we get-

Thus, the magnitude of the gravitational force is.

e) From Newton’s gravitational law, the vector gravitational force exerted onis expressed as:

Here,is the unit vector exerted on.

Substituting the values in the above equation, we get-

Thus, the (vector) gravitational force exerted onis.

f) According to Coulomb’s law, the magnitude of the electric force can be expressed as:

Here, F is the magnitude of the electric force, k is the Coulomb’s constant that is about,are the charges of the masses, andis the distance amongst them.

Substituting the values in the above equation, we get-

Thus, the magnitude of the electric force exerted onis.

g) From Coulomb’s law, the vector electric force exerted onis expressed as:

Here,is the unit vector exerted on.

Substituting the values in the above equation, we get-

Thus, the (vector) electric force exerted onis.

h) The ratio of the magnitude of the electric force to the magnitude of the gravitational force can be expressed as:

Thus, the ratio of the magnitude of the electric force to the magnitude of the gravitational force is.

i) If the distance between the masses becomes, then the distance between the masses can be expressed from equation:

Per equationand for the distanceand charges, the magnitude of the electric force will be:

From equation, for the distance, the magnitude of the gravitational force can be calculated as:

Then, the ratio of the magnitude of the electric force and the gravitational force is-

Thus, the ratio of the magnitude of the electric force and the gravitational force is.