Coulomb's Law/Statics?

Question

Two small spheres of 15 g each are suspended from a common point by threads of length 35 cm. Each thread makes an angle with the vertical of 20 degrees. Each sphere carries the same charge. Find the magnitude of this charge. 

Answer 1

Mass of each sphere is 15 g .

Length of thread is 35 cm .

Each thread makes an angle with the vertical of 20° .

 

For each sphere, a free body diagram shows 

Resolving forces in x and y directions .

T cos 20 = mg 

T sin 20 = F_e 

From above two F_e = mg tan ϴ .

So the force acting on sphere is F_e = mg tan ϴ .

F_e = (15)(9.8) tan 20

F_e = 53.50 N 

Now calculate for distance between two spheres .

Form the above figure sin 20 = (d/2) / 35 cm 

d/2 = sin 20 * 35 cm

d/2 = (0.342) * 35 cm 

d/2 = 11.97 cm 

d = 23.94 cm .

So the distance between two spheres is 23.94 cm .

Let the charge of sphere is q .

The Charge of the sphere can be evaluated using coulomb's law .

F =  (k q² ) / d²

q² = (F * d²)/k

Where proportionality constant k = 9.0 x 10^9 .

q² = (53.50 * (23.94)²)/9.0 x 10^9 

q² = (53.50 * (23.94)²)/9.0 x 10^9 

q² = 3.4069 * 10^-6

q = 1.845 * 10^-3 coulombs

So the charge of the sphere is 1.845 mC . 

 

Comments

thanks for providing the idea

But you forgot to change alll units to SI units