Electric field 2

A point particle with charge q is placed inside a cube but not at its center. The electric flux through any one side of the cube:

• is zero

• is q/2ε₀

• is q/4ε₀

• is q/6ε₀

Answer explanation

The total flux is $\frac{q}{\epsilon o}$ , the flux through one face will be $\frac{q}{6\epsilon o}$

The figure below shows five Gaussian surfaces. Surrounding a distribution of charges. Which surface has a flux of  $\frac{q}{\epsilon o}$  

• a

• b

• c

• d

Answer explanation

 $\phi=\frac{q}{\epsilon o}$  
Surface a= $\frac{-2q+3q-q+2q}{\epsilon o}=\frac{2q}{\epsilon o}$  
Surface b = $\frac{+3q-2q}{\epsilon o}=\frac{q}{\epsilon o}$  
Surface c= $\frac{2q-2q}{\epsilon o}=0$  
Surface d=  $\frac{-2q}{\epsilon o}$  

The figure below shows a distribution of charges. The flux of the electric field due to these charges through the surface E is _____.

• 0

• q/Eo

• 2q/Eo

• 3q/Eo

Answer explanation

$\phi=\frac{q_{enclosed}}{\epsilon o}$
The charges inside the surface are q+q = 2q

The electric flux through a spherical Gaussian surface of radius R centered about an amount of charge Q is 1200 N/C m². What is the electric flux through a cubic Gaussian surface of side R centered about the same charge Q?

• Less than 1200

• More than 1200

• Equal to 1200

• It cannot be determined with the given parameters

Answer explanation

$\phi=\frac{q}{\epsilon o}$
Both shapes contain the same charge and therefore same flux through them

Consider the four cases shown below. Each one contains either a charge q or a charge 2q. A Gaussian surface surrounds the charged particle in each case. Considering the electric flux through each of the Gaussian surfaces, which of the following comparative statements is correct

• 1=3 less  than 2=4

• 1=3 greater than 2=4

• 1=2 less than 3=4

• 3=4 greater than 2=3

Answer explanation

 $\phi=\frac{q}{\epsilon o}$  
Both surfaces 1 and 3 contain the same charge q therefore the flux through them is  $\frac{q}{\epsilon o}$  
And surfaces 2 and 4 contain 2q each therefore the flux through them is double the flux through 1 and 3  $\frac{2q}{\epsilon o}$  

The electric flux through the planar surface above ( positive unit normal to left ) is:

• Positive

• Negative

• Zero

Answer explanation

$\phi=EA\cos\ \theta$

The electric field due to the positive charge is away from it (left) and the electric field from the negative charge is towards it (left) therefore the net electric field is to the left.
The electric field is in opposite direction to the area vector that is the angle between the electric field E and the area vector A is 180
cos 180 = -1 and therefore the flux is negative

• positive (net outward flux)

• negative (net outward flux).

• positive (net inward flux)

• negative (net inward flux)

• zero

Answer explanation

no charge inside the surface therefore flux is zero

What is the general equation for Electric Field?

• $E=\frac{F}{q_0}$
  

• 
  

  $E=\frac{kq_1q_2}{r^2}$

• $E=F\times q_0$

• $E=\lambda ds$

Determine the electric flux of a flat square having an area of 10m2 is a uniform electric field of 8000 N/C passing perpendicular to it

• $8\times10^5\ \frac{Nm^2}{C}$

• $8\times10^4\ \frac{Nm^2}{C}$

• 0

• $16\times10^5\ \frac{Nm^2}{C}$

Answer explanation

$\phi=EA\cos\theta$
The electric field is perpendicular on the area and therefore parallel to the area vector so the angle is zero
$\phi=8000\times100\ \times\ \cos0=80\ 000\ \frac{Nm^2}{C}$

A plane surface is rotated in a uniform electric field. When is the flux of the electric field through the surface maximum?

• When the surface is perpendicular to the field

• When the surface is parallel to the field

• When the surface is at an angle of 30 with the field

• When the surface is at an angle of 45 with the field

Answer explanation

$\phi=EA\cos\theta$
cos $\theta$ is maximum when the angle is zero that is when the electric field and the area vector are parallel, that is when the field is perpendicular to the area (surface)