86. The equation ∇. j = 0 is known as—

1. 1. Poisson's equation

   
   
   

   2. Laplace equation

   
   
   

   3. Continuity equation

   
   
   

   4. Maxwell equation

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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87. A point charge of +1 nC is placed in a space with a permittivity of 8.85 × 10^–12 F/m as shown in figure. The potential difference VPQ between two points P and Q at distance of 40 mm and 20 mm respectively from the point charge is—
    

1. 1. 0.22 kV

   
   
   

   2. – 225 V

   
   
   

   3. – 2.24 kV

   
   
   

   4. 15 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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88. Given the potential function in free space to be V (x) = (50x² + 50y² + 50z²) volts, the magnitude (in volts/metre) and the direction of the electric field at a point (1, – 1, 1), where the dimensions are in metres, are—

1. 1. 100; (^ i + ^ j + ^ k)

   
   
   

   2. 100/√3; (^ i – ^ j + ^ k)

   
   
   

   3. 100 √3; [(–^ i – ^ j – ^ k)/√3]

   
   
   

   4. 100√3; [(–^ i – ^ j – ^ k)/3]

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   →			^		^		^
   E = −		∂Vx	i +	∂Vy	j +	∂Vz	k		volts/m
   		∂x		∂y		∂z

   
   = −[100xi^{^} + 100yj^{^} + 100zk^{^}]volts/m
   At (1, – 1, 1) ^→E = – 100 [^{^}i – ^{^}j + ^{^}k] volts/m
   Its direction is –^{^}i + ^{^}j – ^{^}k/√3
   and magnitude = 100√3
   

   Correct Option: C

   
   

   →			^		^		^
   E = −		∂Vx	i +	∂Vy	j +	∂Vz	k		volts/m
   		∂x		∂y		∂z

   
   = −[100xi^{^} + 100yj^{^} + 100zk^{^}]volts/m
   At (1, – 1, 1) ^→E = – 100 [^{^}i – ^{^}j + ^{^}k] volts/m
   Its direction is –^{^}i + ^{^}j – ^{^}k/√3
   and magnitude = 100√3
   

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89. The maximum usable frequency of an ionospheric layer at 60º incidence and with 8 MHz critical frequency is—

1. 1. 16 MHz

   
   
   

   2. 16/√3 MHz

   
   
   

   3. 8 MHz

   
   
   

   4. about 6.93 MHz
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   MUF = fc sec θ = 8 sec 60º = 16 MHz

   Correct Option: B

   MUF = fc sec θ = 8 sec 60º = 16 MHz

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90. An antenna in free space receives 2 µW of power when the incident electric field is 20 V/m rms. The effective aperture of the antenna is—

1. 1. 0.005 m²

   
   
   

   2. 0.05 m²

   
   
   

   3. 1.885 m²

   
   
   

   4. 3.77 m²

   
   
   

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1. View Hint View Answer Discuss in Forum

   Pr =	|Erms|2	Ae
   	2η0

   
   Given η₀ = 120 πΩ, E_rms = 20 V/m,
   P_r = 20 × 10^–6 watt
   Now,
   

   Ae =	2 × 10−6 × 2 × 120∏
   	20 × 20 × 10−6

   
   = 3.77 m²

   Correct Option: D

   Pr =	|Erms|2	Ae
   	2η0

   
   Given η₀ = 120 πΩ, E_rms = 20 V/m,
   P_r = 20 × 10^–6 watt
   Now,
   

   Ae =	2 × 10−6 × 2 × 120∏
   	20 × 20 × 10−6

   
   = 3.77 m²

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