(a) What type of charge distribution is inside the surface?
(b) If the radius of the cylinder is 0.66 m and the magnitude of the electric field is 300 N/C, what is the net electric flux through the closed surface? How is the electric flux related to the electric field vector and the normal to the surface? What is the orientation of the electric field relative to the curved surface?
(c) What is the net charge inside the cylinder?
(a) Option d is true.
A negatively charged plane parallel to the end faces of the cylinder.
(b) Radius of cylinder, r=0.66m
Magnitude of electric field, E=300 N/C
We have to find the net flux through the closed surface.
Net electric flux,
\[ \phi = -2 EA = -2E(\pi r^2) \]
\[ \phi = -2 \times 300 \times (3.14 \times (0.66)^2) \]
\[ \phi = -820.67 Nm^2/C \]
(c) Net charge,
\[ Q = \epsilon_0 \times \phi \]
Where \( \epsilon_0 = 8.85 \times 10^{-12} \)
\[ Q = -820.67 \times 8.85 \times 10^{-12} \]
\[ Q = -7.26 \times 10^{-9} C \]
\[ Q = -7.26nC \]
Where \( 1nC = 10^{-9}C \)
Explanation:
Charge Distribution Inside the Surface
The correct option for the charge distribution inside the closed cylindrical surface is a negatively charged plane parallel to the end faces of the cylinder. This means that there is an accumulation of negative charges on a plane that is parallel to the end faces of the cylinder.
Net Electric Flux Calculation
To calculate the net electric flux through the closed surface, we use the formula:
\[ \phi = -2E \times A \]
Where E is the magnitude of the electric field and A is the area of the surface. In this case, the area A is equal to \( \pi r^2 \) where r is the radius of the cylinder.
Substitute the given values:
Radius of cylinder, r = 0.66m
Magnitude of electric field, E = 300 N/C
\[ \phi = -2 \times 300 \times (3.14 \times (0.66)^2) \]
\[ \phi = -820.67 Nm^2/C \]
This negative value indicates that the net electric flux through the closed surface is directed inwards.
Net Charge Inside the Cylinder
To find the net charge inside the cylinder, we use the relation between electric flux and charge:
\[ Q = \epsilon_0 \times \phi \]
Where \( \epsilon_0 \) is the permittivity of free space.
Substitute the calculated value of electric flux:
\[ Q = -820.67 \times 8.85 \times 10^{-12} \]
\[ Q = -7.26 \times 10^{-9} C \]
\[ Q = -7.26nC \]
This negative net charge indicates an accumulation of negative charge inside the cylinder.