 # Two parallel plates have equal and opposite charges. When the space between the Donald Johnson 2021-12-17 Answered

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?

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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 ${\sigma }_{i}$
${\sigma }_{i}=\sigma \left(1-\frac{1}{K}\right)$
$=\left(2.833×{10}^{6}\frac{C}{{m}^{2}}\right)\left(1-\frac{1}{1.28}\right)$
$=0.620×{10}^{-6}\frac{C}{{m}^{2}}$
$=0.620\mu \frac{C}{{m}^{2}}$

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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

We have step-by-step solutions for your answer! 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
$a\right)6.2\cdot {10}^{-7}c/{m}^{2}$
$b\right)K=1.28$

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