A point charge of magnitude qq is at the center of a cube with sides of length L. What is the electric flux Φ through each of the six faces of the cube? What would be the flux Φ1 through a face of the cube if its sides were of length L1?

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Nadine Salcido · Accepted Answer

Step 1  The number of field lines passing through a particular area is called Electric Flux.  Consider a point charge of magnitude q is placed at the center of a cube with sides of length L. The electric flux is given by the formula as follows :  Φ=qξ0  q is the charge enclosed in the Gaussian surface NK ξ0 is the permittivity of free space  Step 2   Part (a).   The electric flux Φ through each of the six faces of the cube will be divided into 6 faces. So, using the above relation we get as follows :  ϕ=q6ξ0  Part (b).   Let the side of the cube is L1. The charge enclosed within the closed surface will remain q. So, the electric flux through the face of the cube will be the same as the previous one.  i.e. ϕ1=q6ξ0  Answer:   (a) The electric flux Φ through each of the six faces of the cube is equal to q6ξ0.  (b) The flux Φ1 through a face of the cube if its sides were of length L1 is equal to q6ξ0

ambarakaq8 · Accepted Answer

Answer:   (A)   Consider a cube of side length L in which a charge of magnitude q is placed at the center. Consider a closed Gaussian surface which is cube whose electric flux is Φ which is given by,  Φ=qξ0  The number of field lines emerging from charge will be divided into six faces. So, the electric flux is given by,  Φ=q6ξ0  Part A   The electric flux through each of the six faces of the cube is Φ=q6ξ0  The electric flux is defined as the number of field lines passing per unit area. The flux passing through any face of the cube is equal to the total flux through the cube divided by six.   (B)   Consider the sides of the cube is L1.  As, the electric flux depends only on the enclosed charge q, the flux through each face would be same as previous part even if the dimension of the cube is changed. Hence, the electric flux through each of the six faces of the cube whose length L1 is Φ1=q6ξ0  Part B   The electric flux through each of the six faces of the cube is Φ1=q6ξ0  As the flux depends upon the charge inside the closed surface, the flux through each face would be same as previous part even if the dimension is changed.

star233 · Accepted Answer

Electric flux is given by: ϕ=qξ0 Part A ⇒ϕ=qξ0 this is the total flux through the cube since a cube has six sides therefore flux through each face is given by: ⇒ϕs=ϕ6=q6ξ0 ⇒ϕs=q6ξ0 Part B if its sides were of length L1 then also the magnitude of the flux will remain same as total charge enclosed is same therefore now flux is given by ⇒ϕs=q6ξ0

Don Sumner · Accepted Answer

Step 1: To solve this problem, let&#039;s start by calculating the electric flux through each of the six faces of the cube when its sides have a length of L.The electric flux (Φ) through a closed surface is given by Gauss&#039;s Law:Φ=qε0where q is the magnitude of the point charge, and ε0 is the electric constant (also known as the permittivity of free space).In this case, the charge is located at the center of the cube, and the cube has six identical faces. Therefore, the flux through each face will be the same.Let&#039;s calculate it:Φ=qε0Step 2: Now, let&#039;s calculate the flux (Φ1) through a face of the cube when its sides have a length of L1.The electric flux through a closed surface is still given by Gauss&#039;s Law:Φ1=qε0However, now we need to consider the new side length L₁.Therefore, the flux (Φ1) through a face of the cube with sides of length L1 would be:Φ1=qε0Remember to substitute the appropriate values for q and ε0 into the equations to get the numerical result.Note: The electric constant ε0 is approximately equal to 8.854×10−12C2/N·m2.

nick1337 · Accepted Answer

The electric flux, Φ, through a closed surface is given by Gauss&#039;s Law:Φ=qε0,where q is the charge enclosed by the surface, and ε0 is the electric constant.In this case, the charge enclosed by each face of the cube is the same, since the charge is at the center. Therefore, the electric flux through each face of the cube, Φ, is given by:Φ=qε0.Now, let&#039;s calculate the electric flux Φ1 through a face of the cube if its sides were of length L1.Similar to the previous case, the charge enclosed by the face of the cube is still q, since the charge remains at the center. Therefore, the electric flux through the face of the cube with sides of length L1, Φ1, is given by:Φ1=qε0.

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