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

`m = 0.041 kg`

Step-by-step 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 = √((mg)^2 + (qE)^2)`

also by division of above two equations we have

`(qE)/(mg) = tan\theta`

`qE = mg tan\theta`

now from above equation again

`T = √((mg)^2 + (mg)^2 tan^2\theta)`

`T = mg sec\theta`

`0.420 = m(9.81) sec 16`

`m = 0.041 kg`