First slide

Electric field

Question

A cube of side b has a charge q at seven of its vertices. The electric field due to this charge distribution at the centre of this cube will be:

Moderate

A

$\frac{4 k \cdot q}{3 b^{2}}$
B

kb/2b²
C

32 kb/b²
D

Zero

Solution

If same charge is placed at all eight vertices then net electric field is zero i.e., vector sum of all eight fields is zero.
${\vec{E}}_{1} + {\vec{E}}_{2} + \vec{E_{3}} + \vec{E_{4}} + \vec{E_{5}} + \vec{E_{6}} + \vec{E_{7}} + \vec{E_{8}} = 0 |\vec{E_{1}} + {\vec{E}}_{2} + {\vec{E}}_{3} + {\vec{E}}_{4} + {\vec{E}}_{5} + {\vec{E}}_{6} + {\vec{E}}_{7}| = |{\vec{E}}_{8}| = \frac{4 k \cdot q}{3 b^{2}}$
As half of the diagonal of the cube is $r = \frac{\sqrt{3} b}{2}$
Hence when seven charges are at vertices then electric field at the center $= \frac{4 k \cdot q}{3 b^{2}}$

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