Question

A voltmeter reads 4V when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between plates for the same configuration, voltmeter reads $2\,V$. What is the dielectric constant of the material?
A. 0.5
B. 2
C. 8
D. 10

Answer 1

Hint: We can use the equation connecting the capacitance and the dielectric constant of the material to solve this problem. First write the equation for air as dielectric then write the equation when dielectric slab is inserted. Then we can use the law of conservation of charge to find the value of the dielectric constant.

Complete answer:
It is given that the reading of a voltmeter when it is connected to a parallel plate capacitor with air as dielectric is $4\,V$.
When air is replaced with dielectric slab the voltmeter reading is $2\,V$.
We need to find the value of the dielectric constant of material.
We know that for a capacitor with dielectric the equation for capacitance is given as
$ \Rightarrow $$C = \dfrac{{{\varepsilon _0}kA}}{d}$
Where, ${\varepsilon _0}$ is the permittivity of free space, $k$ is the dielectric constant of the material inserted between the plates of the capacitor, A is the area and d is the distance between the plates of the capacitor.
Let capacitance be ${C_1}$ when air is used as dielectric. Then we can write ${C_1}$ as
$ \Rightarrow $${C_1} = \dfrac{{{k_1}{\varepsilon _0}A}}{d}$
$ \Rightarrow {C_1} = \dfrac{{{\varepsilon _0}A}}{d}$
Since, the dielectric constant of air, ${k_1} = 1$.
 Voltage when air is used as dielectric is given as, ${V_1} = 4\,V$
Now, let the capacitance when dielectric slab is inserted be ${C_2}$.
The capacitance ${C_2}$ can be written as,
\[{C_2} = \dfrac{{{k_2}{\varepsilon _0}A}}{d}\]
Where, ${k_2}$ is the dielectric constant of the dielectric slab.
Voltage when dielectric slab is inserted is, ${V_2} = 2\,V$
We know that charge of a capacitor is given as the product of capacitance and voltage.
$Q = CV$
According to the law of conservation of charge the final charge will be the same as the initial charge.
The initial charge,
$ \Rightarrow $${Q_1} = {C_1}{V_1}$
Final charge,
$ \Rightarrow $${Q_2} = {C_2}{V_2}$
 Thus, using conservation of charge we can write
$ \Rightarrow $${C_1}{V_1} = {C _2}{V_2}$
On substituting the values, we get
$ \Rightarrow $$\dfrac{{{\varepsilon _0}A}}{d}{V_1} = \dfrac{{{k_2}{\varepsilon _0}A}}{d}{V_2}$
$ \Rightarrow {V_1} = {k_2}{V_2}$
$ \Rightarrow 4 = 2{k_2}$
$\therefore k = 2$
This is the dielectric constant of the material.

So, the correct answer is option B.

Note:
Remember that the charge is the product of voltage and capacitance. We know charge remains constant according to the law of conservation of charge. So, to keep the value constant if voltage is decreasing the capacitance should increase. And since capacitance is directly related to the dielectric constant the value of the dielectric constant should be greater than the initial value.