Question

2) A small sphere of mass m = 6.80 g and charge q1 = 28.1 nC is attached to the end of a string and hangs vertically. A second charge of equal mass and charge q2 = −58.0 nC is located below the first charge a distance d = 2.00 cm below the first charge. (a) Find the tension in the string.

Answer 1

Sol: Information provided in the question:

small sphere of mass m = 6.80 g and charge q1 = 28.1 nC = (28.1 x 10^-9 C )

second charge of equal mass m = 6.80 g and charge q2 = −58.0 nC = (-58.0 x 10^-9 C )

Distance between the two sphere (d) = 2 cm = 2/100 = 0.02 m

=> Arrangement can be shown as :

=> Now the attracting force between opposite charhe can be calculate by Coulomb's law :

that is  

Here   

Then F = 9 x 10⁹ ( 28.1 x 10^-9​​​​​​​ x -58.0 x 10^-9 )/ (0.02²)

F = 0.03667 N (acting in upward direction)

=> Now force due to mass of charge acting downward

F' = mg = (6.8x2/1000)x9.81 = 0.133416 N (Downward direction)

=> Hence Net tension in the string:

T = F' - F = 0.133416 N - 0.03667 N = 0.096746 N

So tension in the string (T) = 0.096746 N