Question

There are two identical small metal spheres with charges 38.9 µC and −27.6399 µC. Thedistance between them is 6 cm. The spheres are placed in contact then set at their original distance. Calculate the magnitude of the force between the two spheres at the final position. The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of N.

Answer 1

Answer:

2683.3N

Explanation:

According to coulombs law which states that "the force of attraction existing between two charge q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance (d) between them. Mathematically |F|= k|q1| |q2| /d² where;

F is the force of attraction between the charges

q1 and q2 are the charges

d is the distance between them

k is the coulombs constant

Given |q1|= 38.9 × 10^-6C and |q2| = 27.6399×10^-6C d = 6cm = 0.06m

k = 8.98755 × 109 Nm² /C²

Substituting the given data's in the equation we have;

|F| = 8.98755 × 10^9×38.9×10^-6×27.6399×10^-6/0.06²

|F| = 9.66/0.06²

|F| = 9.66/0.0036

|F| = 2683.3N

The magnitude of the force will be 2683.3N

Note that the modulus of the charges changes negative value of q2 to positive value. The opposite signs of the charges doesn't affect the final calculation, it only tells the force of attraction or repulsion between the charges. Since they are unlike charges, they will attract each other in the field.