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Physics Lesson 14.6.4 - Electric Flux of a Charged Cube
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Welcome to our Physics lesson on Electric Flux of a Charged Cube, this is the fourth lesson of our suite of physics lessons covering the topic of Electric Flux. Gauss Law, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
Electric Flux of a Charged Cube
Let's consider the electric flux across a cube of side length a when it is inserted inside a uniform electric field E. Assume the electric field lines are perpendicular to one of the cube faces.
The four faces parallel to the field have zero flux as no field line crosses them. This is also clear when considering the electric flux formula, in which the cosine between the field lines and the area vectors is zero as they are perpendicular (cos 90° = 0).
From the two remaining faces, it is easy to see that in the left face there is an inwards flux while in the right face the flux is in the outward direction. Therefore, the total flux is zero as the area vectors in these two faces are opposite. This means for one of them the angle must be taken as zero (outward) while for the other face (inward) we must take the angle as 180°. Therefore, we obtain for the total electric flux flowing through the cube
= |E⃗ | × |A⃗ | × cos 1800 +|E⃗ | × |A⃗ | × cos 00
= |E⃗ | × |A⃗ | × (-1) + |E⃗ | × |A⃗ | × 1
= -|E⃗ | × |A⃗ | + |E⃗ | × |A⃗ |
= 0
In general, when electric field is at angle θ to the area vector, we obtain for the total electric flux flowing through a cube
= |E⃗ | × |A⃗ | × cos θ + |E⃗ | × |A⃗ | × cos (180 - θ)
= 0
Example 4
What is the net electric flux for the triangular prism shown in the figure if the prism is placed inside a 20 V/m uniform electric field whose lines are horizontal?
Solution 4
Electric field lines cross only two of the three rectangular faces of prism, because the bottom face and the two bases are parallel to the electric field lines.
First, we find the missing side of the prism using the Pythagorean Theorem. We have
Thus, c = √25 = 5 cm.
The electric flux crossing the 4 cm × 10 cm face is in the inward direction, so we take it as negative, while that on the 5 cm × 10 cm face is in the outward direction, so we take it as positive.
The area vector of the left face is in the opposite direction of the electric field vector, so the angle between them is 180° (cos 180° = - 1). In addition, the magnitude of area vector on the left side (at the input) is
= 40 cm2
= 4 × 10-3 m2
Therefore, the inward flux is
= 20 V/m × 4 × 10-3 m2 × (-1)
= -8 × 10-2 V × m
= -0.08 V × m
The magnitude of area vector for the right face of prism (output) is
= 50 cm2
= 5 × 10-3 m2
Furthermore, the electric field lines are not in the direction of the area vector but they form an angle θ equal to that of the upper vertex (the sides are respectively perpendicular) as shown in the figure.
We need the cosine of the angle θ to work out the flux, i.e.
= 3 cm/5 cm
= 0.6
Therefore, the outward flux is
= 20 V/m × 5 × 10-3 m2 × 0.6
= 6 × 10-2 V × m
= 0.06 V × m
Hence, the net electric flux flowing through the prism is
= -0.08 V × m + 0.06 V × m
= -0.02 V/m
You have reached the end of Physics lesson 14.6.4 Electric Flux of a Charged Cube. There are 4 lessons in this physics tutorial covering Electric Flux. Gauss Law, you can access all the lessons from this tutorial below.
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