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 30.2 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?

Answer 1

Consider initially Charge are at a distance of "x" m as shown in the figure
Taking force balance in vertical direction
TCos$\theta$ = mg --------(1)
where m is mass of charge
Now in horizontal direction
TSin$\theta$ = F
where F is electrostatic force = kq² /x²
TSin$\theta$ = k(q²/x²) ----------(2)
Now dividing 2 by 1
$tan\theta = \left ( \frac{k}{mg} \right )\left ( \frac{q^{2}}{x^{2}} \right )$
On putting the values
$tan30.2 = \left ( \frac{9*10^{9}}{4*9.81} \right )\left ( \frac{q^{2}}{x^{2}} \right )$
$\left ( \frac{q^{2}}{x^{2}} \right ) = 2.54*10^{-9}$
Now in second case where the distance is "2x"
Now in the vertical direction
T = mg + F
where F is electrostatic force = kq² /(2x)² = (k/4)(q²/x²)
$T = mg + \left ( \frac{k}{4} \right )\left ( \frac{q^{2}}{x^{2}} \right )$
Now using the result of case 1
$T = (4*9.81) + \left ( \frac{9*10^{9}}{4} \right )(2.54*10^{-9})$
T = 44.96 N
hence the tension in the string would be 44.96 N when -ve charge will come under the +ve charge.