Answer to Question #89837 in Electricity and Magnetism for Shivam Nishad

Question

A parallel-plate capacitor has plates with
area of cross-section 100 cm² and separated
by 1.0 cm. A potential difference of 100 V
is applied when no dielectric is present. A
dielectric slab of relative permittivity 5 and
thickness 0.5 cm is introduced between the
plates. Calculate the free charge on the
capacitor plates.

Answer 1

According to the formula of capacity for parallel-plate capacitor:

$$C=(\epsilon*\epsilon_0*S)/d (1)$$

Where

$\epsilon$ - relative permittivity

$\epsilon_0$ - electric constant, $\epsilon_0=8.854*10^{−12} F⋅m^{−1}$

$S$ - area of plates

$d$ - distance between plates

After introducing the slab between the plates, the system could be described as 3 capasitors, connected together in series.

First capasitor capacity, according to (1):

$$C_1=(\epsilon*\epsilon_0*S)/x_1 (2)$$

Where

$x_1$ - distance from first plate to slab.

Second capasitor capacity, according to (1):

$$C_2=(\epsilon_2*\epsilon_0*S)/d_2 (3)$$

Where

$\epsilon_2$ - relative permittivity of a slab.

$d_2$ - thickness of a slab

Third capasitor capacity, according to (1):

​

$$C_3=(\epsilon*\epsilon_0*S)/x_2 (4)$$

Where

$x_2$ - distance from last plate to slab.

According the formula of connectected in deries capacitors:

$$1/C=1/C_1+1/C_2+1/C_3 (5)$$

Using (2), (3) and (4):

$$1/C=x_1/(\epsilon*\epsilon_0*S)+d_2/(\epsilon_2*\epsilon_0*S)+x_2/(\epsilon*\epsilon_0*S) (6)$$

$$\implies$$

$$C=(\epsilon*\epsilon_2*\epsilon_0*S)/(\epsilon_2*(x_1+x_2)+\epsilon*d_2) (7)$$

The distance between the plates do not changes $\implies$ $x_1+x_2=d-d_2 (8)$

Where

$d$ - sictance between plates

Using (7), (8):

$$C=(\epsilon*\epsilon_2*\epsilon_0*S)/(\epsilon_2*(d-d_2)+\epsilon*d_2) (9)$$

Acording the definition of capabitor capacity:

$$C=Q/U \implies Q=C*U (10)$$

Where

$Q$ - charge on the plates

$U$ - potential difference

Using (9), (10)

$$Q=U*(\epsilon*\epsilon_2*\epsilon_0*S)/(\epsilon_2*(d-d_2)+\epsilon*d_2)$$

Using the numbers in SI:

$U=100V$

$\epsilon=1$

$\epsilon_0=8.854*10^{−12} F⋅m^{−1}$

$\epsilon_2=5$

$S=100cm^2=10^{-2}m^2$

$d=1cm=10^{-2}m$

$d_2=0.5cm=0.5*10^{-2}m$

Answer: the free charge on the capacitor plates is $Q=1.61*10^{-9} С$