Electrostatics 5 Question 7

8. Voltage rating of a parallel plate capacitor is 500V. Its dielectric can withstand a maximum electric field of 106V/m. The plate area is 104m2. What is the dielectric constant, if the capacitance is 15pF ?

(Take, ε0=8.86×1012C2/Nm2 )

(Main 2019, 8 April I)

(a) 3.8

(b) 8.5

(c) 4.5

(d) 6.2

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Answer:

Correct Answer: 8. Q0=CVR2R1+R2,α=R1+R2CR1R2 9. (a) Q=CV2(1e2t/3RC)

(b) i2=V2RV6Re2t/3RC,V2R

Solution:

  1. As we know, capacitance of a capacitor filled with dielectric medium,

C=ε0KAd

and potential difference between plates is

E=Vdd=VE

So, by combining both Eqs. (i) and (ii), we get

K=CVε0AE

Given, C=15pF=15×1012F,

V=500V,E=106Vm1,

A=104m2

and ε0=8.85×1012C2N1m2

Substituting the values in Eq. (iii), we get

K=15×1012×5008.85×1012×104×106=8.478.5



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