A spherical Gaussian surface of radius R is situated in space along with both conducting and insulating charged objects. The net electric flux through the Gaussian surface is:______

Answers

Answer 1

Answer:

Ф = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]

Explanation:

Radius of Gaussian surface = R

Charge in the Sphere ( Gaussian surface ) = Q

lets take the radius of the sphere to be equal to radius of the Gaussian surface i.e. R

To determine the net electric flux through the Gaussian surface

we have to apply Gauci law

Ф = 4[tex]\pi r^2 E[/tex]

Ф = [tex]\frac{Q_{enc} |}{e_{0} }[/tex]

    = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]


Related Questions

Quadrilateral ABCD is a rectangle.
If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.

Answers

Hope this helps...........

1. Consider a solid cube of dimensions 1ft x 1ft x 1ft (=0.305m x 0.305m x 0.305m). Its top surface is 10
ft (=3.05 m) below the surface of the water. The density of water is pf=1000 kg/m3.
Consider two cases:
a) The cube is made of cork (pB=160.2 kg/m3)
b) The cube is made of steel (pB=7849 kg/m3)
In what direction does the body tend to move?​

Answers

Answer:

  a) up

  b) down

Explanation:

When the cube is less dense than water, it will tend to float (move upward). When it is more dense, it will sink (move downward).

a) 160.2 kg/m^3 < 1000 kg/m^3. The cube will move up.

__

b) 7849 kg/m^3 > 1000 kg/m^3. The cube will move down.

A differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistances are both 1 kΩ?

Answers

Answer:

required feedback resistance ( R2 ) = 100 k Ω

Explanation:

Given data :

Voltage gain = 100

input resistance ( R1 ) = 1 k ohms

calculate feedback resistance required

voltage gain of differential amplifier

[tex]\frac{Vout}{V2 - V1 } = \frac{R2}{R1}[/tex]

= Voltage gain =  R2/R1

= 100 = R2/1

hence required feedback resistance ( R2 ) = 100 k Ω

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