Question

If the electric E field is given by (4î +3j +12k), calaulate the electric flux through
a surface area of 40 units lying in the y-z plane.

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Answer 1

Answer:

Electric flux = 160 N/C units

Explanation:

Given:

• Electric field = 4i + 3j + 12k
• Surface area = 40 units (direction = yz plane)

To Find:

• Electric flux through the surface

Solution:

Here given that the surface area lies in the yz plane

Hence,

$\tt \: Surface\:area=40\: \hat{i}$

Now electric flux is given by the equation,

$\boxed{\tt Electric\:flux \: (\phi)=E.S}$

where E is the electric field and S is the area vector

Substituting the given data we get,

$\sf \phi = (4\hat{i}+3\hat{j}+12\hat{k}).(40\hat{i})$

$\sf \phi=(4 \times 40+3\times0+12\times 0)$

Solving it we get,

$\tt\phi=4\times40$

$\sf \implies \phi=160\: NC^{-1}\: units$

Hence the electric flux through the surface is 160 N/C units.

Answer 2

Answer:

Electric flux = 160 N/C units

Explanation:

Given:

• Electric field = 4i + 3j + 12k
• Surface area = 40 units (direction = yz plane)

To Find:

• Electric flux through the surface

Solution:

• Here given that the surface area lies in the yz plane

Hence,

$\tt \: Surface\:area=40\: \hat{i}$

Now electric flux is given by the equation,

$\boxed{\tt Electric\:flux \: (\phi)=E.S}$

• where E is the electric field and S is the area vector

Substituting the given data we get,

$\sf \phi = (4\hat{i}+3\hat{j}+12\hat{k}).(40\hat{i})$

$\tt \phi=(4 \times 40+3\times0+12\times 0$

Solving it we get,

$\tt\phi=4\times40ϕ=4×40$

$\sf \implies \phi=160\: NC^{-1}\: units$

• Hence the electric flux through the surface is 160 N/C units.