# Two parallel plates have equal and opposite charges. When the space between the

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is $E=3.20×{10}^{5}\mathrm{V}/\mathrm{m}$. When the space is filled with dielectric, the electric field is $E=2.50×{10}^{5}\mathrm{V}/\mathrm{m}$. (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Tiefdruckot

Step 1
Given
We are given the electric field between the two plates ${E}_{0}=3.20×{10}^{5}\frac{V}{m}$ the spase between the two electrodes is vacuum. And the electric field with a dielectric material is $E=2.50×{10}^{5}$ V/m
Reqired
(a) We are asked to finde the charge density ${\sigma }_{i}$ on each surface of the dielectric.
(b) We want to determine the dielectric constant $K$
Solution
(a) When a dielectric material is inserted between the plates while the charge is kept constant with a surface charge density $\sigma$, an induced charge of the opposite sign appears on each surface of the dielectric with a charge density ${\sigma }_{i}$. The charge density ${\sigma }_{i}$ on the dielectric material is related to the charge density $\sigma$ on the electrodes by equatoin 24.16 in the from
${\sigma }_{i}=\sigma \left(1-\frac{1}{K}\right)$
The potential difference between the plates decreases by a factor K. Therefore, the electric field between the plates decreases by the same factor and the dielectric constant is related to the electric field ${E}_{0}$ of the vacuum and the electric field of the dielectric $E$ by equation 24.14
$K=\frac{{E}_{0}}{E}$
$=\frac{3.20×{10}^{5}\frac{V}{m}}{2.50×{10}^{5}\frac{V}{m}}$
=1.28
Where K is a unitless and the cahrge on the plates is constant
Step 2
Also, the charge density $\sigma$ before the dielectric material is related to the applied electric field between two parallel plates by
$\sigma ={ϵ}_{0}{E}_{0}$
$=\left(8.854×{10}^{-12}\frac{{C}^{2}}{N}\cdot {m}^{2}\right)\left(3.20×{10}^{5\frac{V}{m}}\right)$
$=2.833×{10}^{-6}\frac{C}{{m}^{2}}$
Now we can plud our values for into equation (1) to get the surface charge density on the dielectric ${\sigma }_{i}$
${\sigma }_{i}=\sigma \left(1-\frac{1}{K}\right)$
$=\left(2.833×{10}^{6}\frac{C}{{m}^{2}}\right)\left(1-\frac{1}{1.28}\right)$
$=0.620×{10}^{-6}\frac{C}{{m}^{2}}$
$=0.620\mu \frac{C}{{m}^{2}}$

Jimmy Macias
Step 3
(b) The dielectric constant was calculated in part (a) by knowing the electric field after the dielectric and the electric field before the dielectric and equals
K=1.28
The dielectri constant decreases the potential difference and the electric field, while the capacitance increases.
Result
(a)The surface charge density on the dielectric is $\sigma =0.620\mu \frac{C}{{m}^{2}}$.
(b)K=1.28

Don Sumner

Step 1
$C=\frac{Q}{{V}_{ab}}={ϵ}_{0}\frac{A}{d}$
${V}_{ab}=Ed$
$E=\frac{Q}{{ϵ}_{0}A}$
$K=\frac{E}{{E}_{0}}⇒$
(K is the dielectric constant, E is the electric field)
Step 2
$b\right)K=\frac{E}{{E}_{0}}=\frac{3.2}{2.5}=1.28,$
$a\right)\sigma ={ϵ}_{0}{E}_{0}=8.854\cdot {10}^{-12}\cdot 3.2\cdot {10}^{5}=2.833\cdot {10}^{-6}c/{m}^{2}$
${\sigma }_{i}=\sigma \left(1-\frac{1}{K}\right)⇒\left(\text{induced cgarge density on surface of dielctric}\right)$
$=\left(2.883\cdot {10}^{-6}\right)c/{m}^{2}\right)\left(1-\frac{1}{1.28}\right)=6.2\cdot {10}^{-7}c/{m}^{2}$
Result
$a\right)6.2\cdot {10}^{-7}c/{m}^{2}$
$b\right)K=1.28$