Question

Consider the situation shown in figure below. The width of each plate is $b=3\text{m}$. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf $E=1000\text{V}$. All surfaces are frictionless. Calculate the value of $M$ (in $\text{kg}$) for which the dielectric slab will stay in equilibrium.
(Given: $d=1\text{m}$)

• A. $1327.5\times {10}^{-6}$
• B. $2655\times {10}^{-6}$
• C. $2655\times {10}^{-12}$
• D. $1327.5\times {10}^{-12}$

Answer 1

The correct option is A $1327.5\times {10}^{-6}$

We know that,

The electric field attracts the dielectric into the capacitor with a force$$F=\frac{E^{2}b{\epsilon }_0(K-1)}{2d}$$Where,
$b\to$ width of the plates.
$K\to$ Dielectric constant of the material.
$d\to$ Separation between the plates.
$E\to$ EMF.

From the data given in the question, surface is frictionless.

Thus, Force acting on dielectric is balanced by the weight.

So, $\frac{E^{2}b{\epsilon }_0(K-1)}{2d}=Mg$

$\Rightarrow M=\frac{E^{2}b{\epsilon }_0(K-1)}{2dg}$

Substituting the data given in the question we get,

$M=\frac{(1000{)}^2\times 3\times 8.85\times {10}^{-12}\times 1000}{2\times 1\times 10}$

$\Rightarrow M=\frac{{10}^6\times 2655\times {10}^{-12}}{2}$

$\Rightarrow M=1327.5\times {10}^{-6}\text{kg}$

Hence, option (a) is the correct answer.