Question

The electric field between the plates of a parallel-plate capacitor is horizontal, uniform, and has a magnitude E. Asmall object with a charge of -2.16 μC is attached to the string. The tension in the string is 0.420 N , and the angle it makes with the vertical is 16 ∘.What is the mass of the object?

Answer 1

Answer:

$m = 0.041 kg$

Explanation:

As we know that the small particle is in equilibrium at an angle of 16 degree with the vertical

so here we can use force balance in vertical and horizontal direction

$T cos\theta = mg$

$T sin\theta = qE$

now from above equation we have

$T = \sqrt{(mg)^2 + (qE)^2}$

also by division of above two equations we have

$\frac{qE}{mg} = tan\theta$

$qE = mg tan\theta$

now from above equation again

$T = \sqrt{(mg)^2 + (mg)^2 tan^2\theta}$

$T = mg sec\theta$

$0.420 = m(9.81) sec 16$

$m = 0.041 kg$