Q. 133.6( 15 Votes )

# A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Answer :

Given, side of the cube = b units

Charge at each vertex = q C

In the cube shown above,

d = diagonal of any face = = b units

l = diagonal of whole cube = = √3 b units

Thus, the distance of centre of the cube from each vertices = = units

∴ Electric potential at the centre of the cube, V =

(∵ 8 charges)

Where,

q = charge

d = distance from origin=

ϵ_{0} = permittivity of space

∴ V = N units C^{-1}

As the charges as symmetrically placed along the vertex of the cube, the net Electric field will be zero at the centre as the field due to 2 charges placed opposite to each other will cancel, so for all charges, net field will be 0.

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PREVIOUSA charge of 8 m C is located at the origin. Calculate the work done in taking a small charge of –2 × 10–9 C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).NEXTTwo tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:(a) at the mid-point of the line joining the two charges, and(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

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