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Question: Given a uniform electric field \(\vec E = 5 \times {10^3}\hat i\,N{C^{ - 1}}\), find the flux of thi...

Given a uniform electric field E⃗=5×103i^ NC−1\vec E = 5 \times {10^3}\hat i\,N{C^{ - 1}}, find the flux of this field through a square of 10 cm10\,cm on a side whose plane is parallel to the YZYZ plane. What would be the flux through the same square if the plane makes a 30∘{30^ \circ } angle with the XX axis?

Explanation

Solution

Hint
The flux of the electric field is determined by the electric flux on the electric field formula. Initially the electric field parallel to the YZYZ plane, so the angle of the electric flux is taken as zero, then the electric flux can be determined. Then the same electric field is 30∘{30^ \circ } with the XX axis, then the opposite angle of 60∘{60^ \circ } is taken, then the flux can be determined.
The electric flux in the electric field is given by,
⇒F=EScos⁡θ\Rightarrow F = ES\cos \theta
Where, FF is the electric flux, EE is the electric field, SS is the area of the square and θ\theta is the angle of the square.

Complete step by step answer
Given that, The electric field is, E⃗=5×103i^ NC−1\vec E = 5 \times {10^3}\hat i\,N{C^{ - 1}},
The length of the square is, l=10 cml = 10\,cm,
The angle of the square with the XX axis, θ=30∘\theta = {30^ \circ }
Now, The electric flux in the electric field is given by,
⇒F=EScos⁡θ .................(1)\Rightarrow F = ES\cos \theta \,.................\left( 1 \right)
Initially the electric field parallel to the YZYZ plane, so the angle of the electric flux is taken as zero. θ=0∘\theta = {0^ \circ }
By substituting the electric field and the area of the square and the angle in the above equation, then
⇒F=5×103×(10×10−2)2cos⁡0∘\Rightarrow F = 5 \times {10^3} \times {\left( {10 \times {{10}^{ - 2}}} \right)^2}\cos {0^ \circ }
From the trigonometry, the value of the cos⁡0∘=1\cos {0^ \circ } = 1, then
⇒F=5×103×(10×10−2)2\Rightarrow F = 5 \times {10^3} \times {\left( {10 \times {{10}^{ - 2}}} \right)^2}
By squaring the terms in the above equation, then
⇒F=5×103×100×10−4\Rightarrow F = 5 \times {10^3} \times 100 \times {10^{ - 4}}
By multiplying the terms in the above equation, then
⇒F=50 NC−1m2\Rightarrow F = 50\,N{C^{ - 1}}{m^2}
Now the square makes an 30∘{30^ \circ } angle with the XX axis, then the angle is θ=60∘\theta = {60^ \circ }
Substitute the electric field and the area of the square and the angle in the above equation (1), then
⇒F=5×103×(10×10−2)2cos⁡60∘\Rightarrow F = 5 \times {10^3} \times {\left( {10 \times {{10}^{ - 2}}} \right)^2}\cos {60^ \circ }
From the trigonometry, the value of the cos⁡60∘=12\cos {60^ \circ } = \dfrac{1}{2}, then
⇒F=5×103×(10×10−2)2×12\Rightarrow F = 5 \times {10^3} \times {\left( {10 \times {{10}^{ - 2}}} \right)^2} \times \dfrac{1}{2}
By squaring the terms in the above equation, then
⇒F=5×103×100×10−4×12\Rightarrow F = 5 \times {10^3} \times 100 \times {10^{ - 4}} \times \dfrac{1}{2}
By multiplying the terms in the above equation, then
⇒F=25 NC−1m2\Rightarrow F = 25\,N{C^{ - 1}}{m^2}
Thus, the electric flux when the square is parallel to the YZYZ plane is, F=50 NC−1m2F = 50\,N{C^{ - 1}}{m^2}.
Thus, the electric flux when the square is 30∘{30^ \circ } angle with the XX axis is, F=50 NC−1m2F = 50\,N{C^{ - 1}}{m^2}.

Note
The electric flux is directly proportional to the electric field in the square and the area of the square and the angle of the square. If the electric field in the square and the area of the square increases, the electric flux in the square increases.