The electric field between two parallel plates of a capacitor is 2.1 × 10-5 . If a medium is inserted between the plates than the electric field becomes 1.0 × 10-5 . Now, the value of dielectric will be

Question

The electric field between two parallel plates of a capacitor is 2.1 × 10-5 . If a medium is inserted between the plates than the electric field becomes 1.0 × 10-5 . Now, the value of dielectric will be (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

E=(σ /ε 0)=2.1× 10-5 E=(σ /ε 0K)=1× 10-5 ∴ (E/E)=(σ /ε 0)× (ε 0K/σ )=(2.1× 10-5/1× 10-5) or K=2.1≈ 2

Q. The electric field between two parallel plates of a capacitor is $2.1 \times 10^{- 5}$ . If a medium is inserted between the plates than the electric field becomes $1.0 \times 10^{- 5}$ . Now, the value of dielectric will be

2

3

4

5

$E = \frac{σ}{ε _{0}} = 2.1 \times 10^{- 5}$ $E = \frac{σ}{ε _{0} K} = 1 \times 10^{- 5}$ $∴$ $\frac{E}{E} = \frac{σ}{ε _{0}} \times \frac{ε _{0} K}{σ} = \frac{2.1 \times 10 ^{- 5}}{1 \times 10 ^{- 5}}$ or $K = 2.1 \approx 2$

Q. The electric field between two parallel plates of a capacitor is 2.1×10−5 . If a medium is inserted between the plates than the electric field becomes 1.0×10−5 . Now, the value of dielectric will be

2

3

4

5

E=ε0​σ​=2.1×10−5 E=ε0​Kσ​=1×10−5 ∴ EE​=ε0​σ​×σε0​K​=1×10−52.1×10−5​ or K=2.1≈2

Q. The electric field between two parallel plates of a capacitor is $2.1 \times 10^{- 5}$ . If a medium is inserted between the plates than the electric field becomes $1.0 \times 10^{- 5}$ . Now, the value of dielectric will be

$E = \frac{σ}{ε _{0}} = 2.1 \times 10^{- 5}$ $E = \frac{σ}{ε _{0} K} = 1 \times 10^{- 5}$ $∴$ $\frac{E}{E} = \frac{σ}{ε _{0}} \times \frac{ε _{0} K}{σ} = \frac{2.1 \times 10 ^{- 5}}{1 \times 10 ^{- 5}}$ or $K = 2.1 \approx 2$