Question

A positive charge with a mass of 4 kg is placed on a string and a negative charge of the same amount (magnitude) but negative is placed directly to the right of this charge when in equilibrium. Also, the angle with the vertical of the string is 43.31 degrees when the charge is in equilibrium. Then the negative charge is moved directly below the charge on the string and it is placed twice as far as it originally was to the charge on the string. What is the amount of tension in the string now in Newtons? A-q

Answer 1

Initially :

The free body diagram for the given situation can be shown as

In the diagram,

mg = 4 x 9.8 = 39.2 N is force due to gravity

T is tension in the string

+q and -q are two charges

is electrostatic force

and r is distance between the charges

From the free body diagram, we can write

   .........................(1)

cOS

and  

or,   .......................(2)

Tsine = 4,0

Dividing equation (2) by equation (1), we get

.......................(3)

Given,

We have,

Putting these values in the equation (3), we get

or,

or,    ...........................(4)

Final situation :

Now the negative charge is placed directly below the charge twice as far as it originally was.

The free body diagram will look like

Where, T_new is the tension in the string in this situation

F_new is new electrostatic force between two charges

And, F_new can be calculated as

or,

or, (used equation (4))

or,   

Now, from the free body diagram we can write

or,

Tnew = 9.238 + 39.2 48.438 M

Hence, the tension in the string in this new situation will be 48.438 N .

For any doubt please comment.