Question

Physics Question

Topic:- Relative permittivity.

Two identical charged spheres are suspended by strings of equal length. The strings make an angle of 60° with each other . When suspended in a liquid of density 0.5g/cc. The angle remains the same. The dielectric constant of the liquid is? [density of the material of the sphere is1.5g/cc]

Answer with step by step process.
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Answer 1

Hello!!

let, in vacuum figure (1)

θ = 30°

Tsinθ = Fo

Tcosθ = mg

Fo = Force in a vacuum

dividing both we get,

$\tan(θ) \: \: \frac{fo}{mg} \: \: = \: \: \frac{fo}{ \alpha vg} \: \: - - (1) \\ \\ \alpha \: \: = \: \: density \: \: of \: \: sphere \\ \\ v \: \: = \: \: volume \: \: of \: \: sphere \\ \\ m \: \: = \: \: mass \: \: of \: \: sphere \\ \\ m \: \: = \: \: \alpha v$

In Liquid, figure(2)

θ= 30°

T'Sinθ = Fm

T'Cosθ = mg - U

Fm = Force in a liquid medium.

U = upthrust force applied by water on the sphere.

Divide both,

$\tan(θ) \: \: = \: \: \frac{fm}{ mg - u} \: \: = \: \: \frac{fm}{ \alpha vg - v \beta g} \: \: \: - - (2) \\ \\ \beta \: \: = \: \: density \: \: of \: \: liquid \\ \\ v \: \: = \: \: volume \: \: of \: \: sphere \\ \\ u \: \: = \: \: \beta vg$

equating Eq.(1) and Eq.(2)

$\frac{fo}{ \alpha vg} \: \: = \: \: \frac{fm}{ \alpha vg \: - \: v \beta g} \\ \\ \frac{fo}{ \alpha } \: \: = \: \: \frac{fm}{ \alpha - \beta } \\ \\ \frac{fo}{1.5} \: \: = \: \: \frac{fm}{1.5 - 0.5} \\ \\ fm \: \: = \: \: \frac{fo}{1.5}$

fo/fm = Er

dielectric constant [ Er ] = 1.5

Thanks!!

₳₳

Answer 2

Answer:

hope it helps u ☑️☑️☑️☑️

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