1) A total charge of 4.03 C is distributed on two metal spheres. When the sphere

Question

1) A total charge of 4.03 C is distributed on two metal spheres. When the spheres are 10.00 cm apart, they each feel a repulsive force of 3.3*10^11 N. How much charge is on the sphere which has the lower amount of charge?

2) 2 equal charges, 20 micro Coulomb each, are separated by 5 cm. Find force between those

3)Add 7 micro Coulombs and 6,686 nano Coulombs and get the result in nano Coulombs unit.

4)How many excess electrons can be counted in a rubber rod of - 7 *0.00001 nano Coulombs of charges?

5)A metal ball has - 1 Coulomb of charge. If it receives 0.50*10^20 number of electrons, what will be the resultant charge of the ball?

6)2 equal charges are separated by 42 mm and experiences a repulsion force of 9 Newton. Calculate the amount of each charge in micro Coulombs.

7)2 equal charges, 108 micro Coulombs each, experiences an attraction force of 108 Newton. Calculate the distance between the charges in mm

8)What is the total charge of all protons in 9 gram of water (H2O)?

9)3 charges, 2 µC each, are located on three vertices A, B, C of an equilateral triangle with sides 1 cm each. Another charge q is located at the mid point D of the side BC. Calculate q in micro Coulomb so that net force on the charge at A due to the charges at B, C and D is zero

10)In a right angle traingle ABC, angle ABC is 90 Degree, AB = 2 m, and angle ACB is 41.81 Degree. A point charge of 5*47 nC is placed at point C, point charge 4* 47 nC is placed at point A and point charge 1 C is placed in point B. Calculate the force on charge at B due to others two

11) What is the total charge of all electrons in 16 /100 gram of Carbon dioxide CO2 gas?

12) Two equal charges with magnitude Q and Q experience a force F1 when held at a distance r. F2 is the force between two charges of magnitude 7*Q and 2*Q when held at a distance r/10? What is the ratio of F2 to F1?

Explanation / Answer

Force on one sphere due to the other = 9e9*q1*q2/r^2
3.3e11*0.1^2 = 9e9*q1*q2
q1*q2 = 0.367 C^2
q1+ q2 = 4.03 C

q^2 - 4.03 q + 0.367 = 0

Finding the roots we have
q1 = 3.94 C
q2 = 0.093 C

2) Use formula, Force, F = (9e9 X 20 X 20 X 10e-12)/(0.05^2) = 1440 N

3) 0.13686 nC

4) q = ne

n = (7 X 0.00001)/(1.6 X 10-19) = 4.375 X 1014

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