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The total flux through the faces of the ...

The total flux through the faces of the cube with side of length a if a charge q is placed at corner A of the cube is

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`(q)/(6 epsilon_(0))` through each face.
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What will be the total flux through the faces of the cube, with side of length a if a charge q is placed at (a) A : a corner of the cube (b) B : mid-point of an edge of the cube (c ) C : center of a face of the cube (d) D : mid-point of B and C.

A point charge q is placed at the centre of a cube. What is the flux linked. a with all the faces of the cube? b. with each face of the cube? c. if charge is not at the centre, then what will be the answer of parts a and b ?

A point charge q is placed at the centre of a cube. What is the flux linked. a with all the faces of the cube? b. with each face of the cube? c. if charge is not at the centre, then what will be the answer of parts a and b ?

What is the flux through a cube of side 'a' if a point charge of q is at one of its corner :

If identical charges (-q) are placed at each corner of a cube of side b , then electric potential energy of charge (+q) which is placed at centre of the cube will be

The length of each side of a cubical closed surface is l If charge q is situated on one of the vertices of the cube , then find the electric flux passing through shaded face of the cube. .

Consider a point charge q=1 m C placed at a corner of a cube of sides 10 cm . Determine the electric flux through each face of the cube. Strategy : We will learn about the utility of symmetry in solving problems with the help of Gauss's law . Here we'll use the symmetry of the situation,which involves the faces joining at the corner at which the charge resides. (a) A charge q is placed atthe corner of a cube. (b) By surrounding the charge with a series of cubes such that the charge is at the centre of a larger cube, we have created an arrangement sufficiently symmetric to be able to solve for desired flux values. You can see from figure that for these faces vec(E).hat(n)=0 . Since the normal is perpendicular to the surfaces while the electric field goes off in a spherically symmetric pattern and lies in the sides . In other words , the electric field that originates at the charge is tangential to the surface of these three sides. This means there is no flux through these sides. The electric flux through each of the remaining three faces of the be must be equal by symmetry . We'll referto these faces with the label F. To fidn the flux through each of the sides F, we acan use a technique that puts the single charge in the middle of a larger cube. It takes seven other similarly placed cubes to surrounds the points charge q completely in figre. The charge is at the dead centre of the enw larger cube. So, the flux through each of the six sides of the large cube will now have an electric flux of one sixth of the total flux F. So given that the total structure is completely symmetric, the flux through a side F is one fourth of the flux through the larger side.

The length of each side of a cubical closed surface is L metre. If charge 48C is situated at one of the corners of the cube, Find the flux passing through the cube. (In Volt-metre)

the total flux of electric lines of forces of a charge q placed at the mid point of the dge of a side of rectangular box, as shown in figure through the box is

Choose the correct option: Consider a cube of side 'a' as shown. Eight point charges are placed at the corners. The cube is rotated about the central axis with constant angular velocity omega :

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