Two tiny spheres of mass 5.80 mg carry charges of equal magnitude, 77.0 nc , but opposite sign. they are tied to the same ceiling hook by light strings of length 0.530 m. when a horizontal uniform electric field e that is directed to the left is turned on, the spheres hang at rest with the angle θ between the strings equal to 58.0∘ what is the magnitude e of the field?

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

he made 11 calls in one day

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