Respuesta :
Mass m = 5.80 mg = 5.80 x 10⁻⁶ kg
Charge q = 77 nC = 77 x 10⁻⁹ C
String length L = 0.530 m
angle θ = 58.0°
Here we can force on one mass
gravitational force F1= mg is acting in downward direction
F1 = (5.80 x 10⁻⁶ kg)(9.81 m/s²)
F1 = 57 x 10⁻⁶ N
Second force is electrostatic force due to another charge
F2 = kqq/r₂
Where r is the distance between two charges
r = 2 LSinθ = 2(0.53)Sin58
r = 0.9 m
F2 = (9x 10⁹)(77 x 10⁻⁹)(77 x 10⁻⁹)/(0.9²)
F2 = 65.9 x 10⁻⁶ N
Third force is F3 = qE
Fourth force is tension in the string
In equilibrium sum of forces along horizontal will be zero
and sum of forces along vertical will be zero
First we will use sum of vertical forces equal to zero
TCosθ - F1 =0
T Cos58 - ( 57 x 10⁻⁶) = 0
T = 107.56 x 10⁻⁶ N
Now Sum of horizontal forces = 0
F3 - F2 - TSinθ = 0
qE - (65.9 x 10⁻⁶ ) - (107.56 x 10⁻⁶) = 0
(77 x 10⁻⁹) E = 173.46 x 10⁻⁶
E = 2253 N/C
