156. The electric field on the surface of a perfect conductor is 2 V/m. The conductor is immersed in water with ε = 80 ε₀. The surface charge density on the conductor is—

1. 1. 0 c/m²

   
   
   

   2. 2 c/m²

   
   
   

   3. 1.8 × 10^{– 11} c/m²

   
   
   

   4. 1.41 × 10^{– 9} c/m²

   
   
   

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1. View Hint View Answer Discuss in Forum

   Electric field is given by
   

   E =	σ
   	ε0

   
   for the system to be immersed in water its dielectric changes form ε0 to 80 ε0 so we get surface charge density as
   

   σ = ε0εr|→E| ⇒ 80ε0 E =	80 × 10–9	× 2
   	36π

   
   = 1.41 × 10^–9.

   Correct Option: D

   Electric field is given by
   

   E =	σ
   	ε0

   
   for the system to be immersed in water its dielectric changes form ε0 to 80 ε0 so we get surface charge density as
   

   σ = ε0εr|→E| ⇒ 80ε0 E =	80 × 10–9	× 2
   	36π

   
   = 1.41 × 10^–9.

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157. Distilled water at 25ºC is characterized by σ = 1.7 × 10^{– 4} mho/m and ε = 78 ε₀ at a frequency of 3 GHz, its loss tangent, tan δ is—

1. 1. 1.3 × 10^{– 5}

   
   
   

   2. 1.3 × 10^{– 3}

   
   
   

   3. 1.7 × 10^{– 4}/ 78 ε = 10^{– 9}/ (36 π) F/m

   
   
   

   4. 1.7 × 10^{– 4}/ 78 ε₀
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Loss tangent is given by
   

   tan φ =	σ	=	1.7 × 10–4
   	ωσ		2π × 3 × 109 × 78 × 8.6 × 10–12

   
   

   =	1·7 × 10–4 × 36π	= 1.3 × 10–5
   	2π × 3 × 109 × 10-9 × 78

   Correct Option: A

   Loss tangent is given by
   

   tan φ =	σ	=	1.7 × 10–4
   	ωσ		2π × 3 × 109 × 78 × 8.6 × 10–12

   
   

   =	1·7 × 10–4 × 36π	= 1.3 × 10–5
   	2π × 3 × 109 × 10-9 × 78

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158. A plane wave is characterised by this wave is—
     
     

     →		^	^
     E =	(0.5	x +	x	Ejπ/2)	ejωt– jk2

1. 1. Linearly polarized

   
   
   

   2. Circularly polarized

   
   
   

   3. Elliptically polarized

   
   
   

   4. Unpolarized

   
   
   

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1. View Hint View Answer Discuss in Forum

   Wave polarization may be found as
   

   φ = tan–1		Ey		= tan–1	ejπ/2· ejωt – jkz
   		Ex			0.5 · ejωt – jkz

   
   

   = tan–1		j sin π/2		= tan–1		j		= tan–1(2j)
   		0.5				.5

   
   φ = multiple of π/2 so it is circularly polarized.

   Correct Option: B

   Wave polarization may be found as
   

   φ = tan–1		Ey		= tan–1	ejπ/2· ejωt – jkz
   		Ex			0.5 · ejωt – jkz

   
   

   = tan–1		j sin π/2		= tan–1		j		= tan–1(2j)
   		0.5				.5

   
   φ = multiple of π/2 so it is circularly polarized.

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159. The phase velocity for the TE₁₀ mode in an air filled rectangular waveguide is–

1. 1. Less than C

   
   
   

   2. Equal to C

   
   
   

   3. Greater than C

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Always greater than C since the value of sin θ is always less than 1. So 1/sin θ is always greater than 1 that's why V_P > V_C.

   Correct Option: C

   Always greater than C since the value of sin θ is always less than 1. So 1/sin θ is always greater than 1 that's why V_P > V_C.

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160. In an impedance smith chart a clockwise movement along a constant resistance circle gives rise to—

1. 1. A decrease in the value of reactance

   
   
   

   2. An increase in the value of reactance

   
   
   

   3. No change in the reactance value

   
   
   

   4. No change in the impedance value

   
   
   

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1. View Hint View Answer Discuss in Forum

   An increase in the value of reactance.

   Correct Option: B

   An increase in the value of reactance.

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