2025 PHYS 118|PHYS_V 118– Worksheet 3: Gauss’s law WITH WELL EXPLAINED SOLVED ANSWERS SOLUTION
University of British Columbia
Electric flux
Electric flux, Φ𝑒, quantifies the amount of electric field passing through a surface (open or closed). This amount depends
on the area of the surface, on the strength of the field, and on the mutual orientation of the surface and the field.
Qualitatively, you can interpret electric flux as a number of field lines poking through this area. More rigorousely, for a
uniform electric field 𝐸�⃗ and for a plane surface with an area 𝐴, the electric flux is:
Φ𝑒 = 𝐸�⃗ ∙ 𝐴⃗ = 𝐸𝐴 cos 𝜃𝜃 (1)
where 𝐴⃗ is the so-called area vector, i.e. a vector with the magnitude equal to the area 𝐴 of the surface (to account for
the size of the surface) and directed perpendicullary to it (to account for mutual orientation of the field and the surface).
The point “∙” indicates the dot product, and 𝜃𝜃 is the angle between 𝐸�⃗ and 𝐴⃗.
For this surface, draw the normal
unit vector 𝑛�⃗ and the area vector 𝐴⃗:
Note: Units of �𝐴⃗� are m2.
If the electric field is non-uniform, or if the surface is curved, then the flux can be calculated by integration. Basically, you
mentally cut your surface into very small pieces of area 𝑑𝐴, and calculate the flux 𝑑Φ𝑒 through each of them using
equation (1): 𝑑Φ𝑒 = 𝐸�⃗ ∙ 𝑑𝐴⃗. Then the net electric flux is obtained by adding up all these tiny fluxes:
Φ𝑒 = ∫area 𝑑Φ𝑒 = ∫area 𝐸�⃗ ∙ 𝑑𝐴⃗ . (2)
Let’s cut our curved surface into tiny squares. Across each small
square of area Δ𝐴 the electric field is practically constant
(though it might change a lot from square to square), and each
small square is practically flat. Hence, the flux through 𝑛-th
square is ΔΦ𝑛 = 𝐸�⃗ (𝑥 𝑛 , 𝑦𝑛, 𝑧𝑛) ⋅ Δ𝐴⃗𝑛. The total flux through this
surface is approximated by the sum of these tiny fluxes:
Φ𝑒 ≈ ∑𝑛 ΔΦ𝑛 = ∑𝑛 𝐸�⃗ (𝑥 𝑛 , 𝑦𝑛, 𝑧𝑛) ⋅ Δ𝐴⃗𝑛.
To make this expression exact, we have to make the squares
infinitesimally small and change the sum to the integral. This
gives us Eq.(2).
Note: In this integral, 𝐸�⃗ is the electric field is at the location of the tiny area 𝑑𝐴. Hence, if the field is non-uniform, then
𝐸�⃗ depends on the location, and it might be really difficult to calculate this integral analytically. Fortunately, we will never
have to do such a brute-force integration (see below).
, Q3.1. A uniform electric field is passing through a flat rectangular sheet that is lying in the horizontal plane. What is
the flux through the sheet of area A?
EA
E A cos(30)
E A tan(60)
E A cos(60)
Zero
Convention: For a closed surface, we always take the area vector in the outward direction.
Q3.2. Draw the area vectors for
sides A, B, C of the prism shown in
the figure. Your drawing should
reflect both the directions and the
magnitudes of the area vectors.
Q3.3. Prove that for a closed surface:
E-field going ‘outwards’ corresponds to a positive flux
E-field going ‘inwards’ corresponds to a negative flux
University of British Columbia
Electric flux
Electric flux, Φ𝑒, quantifies the amount of electric field passing through a surface (open or closed). This amount depends
on the area of the surface, on the strength of the field, and on the mutual orientation of the surface and the field.
Qualitatively, you can interpret electric flux as a number of field lines poking through this area. More rigorousely, for a
uniform electric field 𝐸�⃗ and for a plane surface with an area 𝐴, the electric flux is:
Φ𝑒 = 𝐸�⃗ ∙ 𝐴⃗ = 𝐸𝐴 cos 𝜃𝜃 (1)
where 𝐴⃗ is the so-called area vector, i.e. a vector with the magnitude equal to the area 𝐴 of the surface (to account for
the size of the surface) and directed perpendicullary to it (to account for mutual orientation of the field and the surface).
The point “∙” indicates the dot product, and 𝜃𝜃 is the angle between 𝐸�⃗ and 𝐴⃗.
For this surface, draw the normal
unit vector 𝑛�⃗ and the area vector 𝐴⃗:
Note: Units of �𝐴⃗� are m2.
If the electric field is non-uniform, or if the surface is curved, then the flux can be calculated by integration. Basically, you
mentally cut your surface into very small pieces of area 𝑑𝐴, and calculate the flux 𝑑Φ𝑒 through each of them using
equation (1): 𝑑Φ𝑒 = 𝐸�⃗ ∙ 𝑑𝐴⃗. Then the net electric flux is obtained by adding up all these tiny fluxes:
Φ𝑒 = ∫area 𝑑Φ𝑒 = ∫area 𝐸�⃗ ∙ 𝑑𝐴⃗ . (2)
Let’s cut our curved surface into tiny squares. Across each small
square of area Δ𝐴 the electric field is practically constant
(though it might change a lot from square to square), and each
small square is practically flat. Hence, the flux through 𝑛-th
square is ΔΦ𝑛 = 𝐸�⃗ (𝑥 𝑛 , 𝑦𝑛, 𝑧𝑛) ⋅ Δ𝐴⃗𝑛. The total flux through this
surface is approximated by the sum of these tiny fluxes:
Φ𝑒 ≈ ∑𝑛 ΔΦ𝑛 = ∑𝑛 𝐸�⃗ (𝑥 𝑛 , 𝑦𝑛, 𝑧𝑛) ⋅ Δ𝐴⃗𝑛.
To make this expression exact, we have to make the squares
infinitesimally small and change the sum to the integral. This
gives us Eq.(2).
Note: In this integral, 𝐸�⃗ is the electric field is at the location of the tiny area 𝑑𝐴. Hence, if the field is non-uniform, then
𝐸�⃗ depends on the location, and it might be really difficult to calculate this integral analytically. Fortunately, we will never
have to do such a brute-force integration (see below).
, Q3.1. A uniform electric field is passing through a flat rectangular sheet that is lying in the horizontal plane. What is
the flux through the sheet of area A?
EA
E A cos(30)
E A tan(60)
E A cos(60)
Zero
Convention: For a closed surface, we always take the area vector in the outward direction.
Q3.2. Draw the area vectors for
sides A, B, C of the prism shown in
the figure. Your drawing should
reflect both the directions and the
magnitudes of the area vectors.
Q3.3. Prove that for a closed surface:
E-field going ‘outwards’ corresponds to a positive flux
E-field going ‘inwards’ corresponds to a negative flux
