2 equal charges, 23 micro Coulomb each, are separated by 5 cm. Find force betwee

Question

2 equal charges, 23 micro Coulomb each, are separated by 5 cm. Find force between theose. How many excess electrons can be counted in a rubber rod -1 *0.00001 nano Coulombs of charges? A metal ball has - 8 Coulomb of charge. If it receives 0.50*10^20 number of electrons, what will be the resultant charge of the ball? 2 equal charges are separated by 47 mm and experiences a repulsion force of 9 Newton. Calculate the amount of each charge in micro Coulombs. 2 equal charges, 191 micro Coulombs each, experiences an attraction force of 191 Newton. Calculated the distance between the charges in mm. What is the total charge of all protons in 8 gram of water (H_2 O)? A total charge of 3.39 C is distributed on two metal spheres. When the spheres are 10.00 cm apart, they each feel a repulsive force of 3.2*10^11 N. How much charge is on the sphere which has the lower amount of charge? In a right angle triangle ABC, angle ABC is 90 Degree, AB = 2 m, and angle ACB is 41.81 Degree. A point charge of 5* 43 nC is placed at point C, point charge 4* 43 nC is placed at point A and point charge 1 C is placed in point B. Calculate the force on charge at C due to others two. Two equal charges with magnitude Q and Q experience a force 5.0050 when held at a distance r. What is the force between two charges of magnitude 2Q and 2Q when held at a distance r/2? How many electrons must be removed from each of two 5.0-kg copper spheres to make the electronic force of repulsion between them equal in magnitude to the gravitational attraction between them?[2.675* 10^9] What is the ratio of the electric force to the gravitational force between a proton and an electron separated by 7.3 times 10^-11 m?

Explanation / Answer

Here ,

1) let the force between them is F

F = k * q^2/d^2

F = 9 *10^9 * (23 *10^-6)^2/(0.05)^2

F = 1904 N

the force between them is 1904 N

2)

number of excess electrons = 0.0001*10^-9/(1.602 *10^-19)

number of excess electrons = 6.24 *10^5

3)for the final charge on the ball

final charge on the ball = initial charge + charge added

final charge on the ball = -8 + 0.50 *10^20 * (-1.602 *10^-19)

final charge on the ball = -16.01 C

the final charge on the ball is -16.01 C

4)

let the each charge is q

9 = 9 *10^9 * q^2/0.047^2

solving for q

q = 1.49 *10^-6 C = 1.49 uC

the final charge is 1.49 uC

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