Two parallel plates have equal and opposite charges- When the space between the plates is evacuated- the electric field is E0-3-20- 105 V-m- When space is filled with a dielectric- the electric field is E-2-50- 105 V-m- The dielectric constant is approximately given as x-1010- Find x-mention answer in decimal

When #160;the space between the plates is evacuated, the electric field is #160; E=QAϵ_0 #160;where A be the area of plates.When the space is filled ...

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Standard XII

Physics

New Capacitance in Dielectric

Two parallel ...

Question

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is $E_{0} = 3.20 \times 10^{5} V / m$ . When space is filled with a dielectric, the electric field is $E = 2.50 \times 10^{5} V / m$ . The dielectric constant is approximately given as $\frac{x + 10}{10}$ . Find $x$ .(mention answer in decimal)

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Solution

When the space between the plates is evacuated, the electric field is $E = \frac{Q}{A ϵ_{0}}$ where A be the area of plates.
When the space is filled with dielectric, the electric field is $E = \frac{Q}{A K ϵ_{0}}$ where K is the dielectric constant.
$\frac{E_{0}}{E} = \frac{\frac{Q}{A ϵ_{0}}}{\frac{Q}{A K ϵ_{0}}} = K$
$∴ K = \frac{E_{0}}{E} = \frac{3.20 \times 10^{5}}{2.50 \times 10^{5}} = 1.28 \approx 1.3 = \frac{13}{10}$

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When the space between the plates is evacuated, the electric field is E=QAϵ0 where A be the area of plates.When the space is filled with dielectric, the electric field is E=QAKϵ0 where K is the dielectric constant.E0E=QAϵ0QAKϵ0=K∴K=E0E=3.20×1052.50×105=1.28≈1.3=1310