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

Solution

Given,
Side of cube=b
Charge on vertices of cube =q
Diagonal DF of cube

DF = \sqrt{b^{2} + b^{2} + b^{2}} DF = b \sqrt{3}

Thus,

DO = \frac{DF}{2} = \frac{\sqrt{3}}{2} b

Due to one charge q the potential at the centre O is given by

V = \frac{1}{4 {πε}_{0}} \frac{q}{r} = (\frac{1}{4 {πε}_{0}} \frac{q}{\frac{\sqrt{3} b}{2}})

Due to eight charges the total potential at the centre O is given as,

V = 8 (\frac{1}{4 {πε}_{0}} \frac{q}{\frac{\sqrt{3}}{2} b})

= \frac{4 q}{\sqrt{3} {πε}_{0} b}

Electric field at the centre of of cube:
Due to two opposite corners D and F electric field intensity at the centre ‘O’ are equal in magnitude and opposite in direction. Therefore, they cancel out each other. Similarly all other intensities cancel out each other and the total electric field at centre is zero.

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Electrostatic Potential And Capacitance

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

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