Question

A positive charge with a mass of 3 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 37.01 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? +q

Answer 1

T = tension force in the string

r_i = distance between the charges = a

m = mass of the charge = 3 kg

= angle of the string with the vertical = 37.01 deg

F_e = electric force between the two charges

electric force between the two charges is given as

F_e = k q²/r_i²

F_e = k q²/a²

From the force diagram , using equilibrium of force in vertical direction , force equation is given as

T Cos = mg eq-1

From the force diagram , using equilibrium of force in horizontal direction , force equation is given as

T Sin = F_e eq-2

dividing eq-2 by eq-1

(T Sin)/(T Cos) = F_e /(mg)

tan = F_e /(mg)

tan37.01 = F_e /(3 x 9.8)

F_e = 22.2 N

Case 2 : , negative charge is placed below the charge on string at twice the distance initially

T' = tension force in the string

r_f = new distance between the charges = 2a

m = mass of the charge = 3 kg

F'_e = new electric force between the two charges

new electric force between the two charges is given as

F'_e = k q²/r_f²

F'_e = k q²/(2a)²

F'_e = k q²/(4a²)

F'_e = F_e/4

From the force diagram , using equilibrium of force in vertical direction , force equation is given as

T' = mg +  F'_e

T' = mg + F_e/4

T' = (3 x 9.8) + 22.2/4

T' = 35 N