Question

10. Six point charges (each of magnitude Q) are
placed on the six vertices of a cube of side x,
such that two adjacent vertices are vacant.
Electrostatic field intensity at the centre of the
cube is k=
ic
K
ATEO
ko
53 x?
412
3 x?
802 KO
33 x?​

Answer 1

Answer:

$\frac{8 \sqrt{2} kq}{3 \sqrt{3} x^2}$

Explanation:

1) We have, 6 point charges of same magnitude 'q' located on the vertices of a cube of side 'x'units .

We try to make the make symmetricity of the cube by including +q &-q charges on the vacant vertices.

2) Forgetting about -q charges, we get symmetric cube with all 8 vertices filled with charge 'q'.

Since, by looking at the centre all charges (+) are symmetric, that is, equidistant in all directions.

Hence, electric field due to 8 point charges (+) is zero.

3) Now, take '-q' charge into the picture.

We know,

Diagonal of cube is given by √3 x.

So, centre to vertex ,a = √3x/2.

Magnitude of Electric field due to -q charge is given by

$E = \frac{kq}{a^2}$

4) We have,

AO =√3x/2,

BO =√3x/2

AB = x

By cosine rule,

$cos(\angle AOB) = \frac{AO^2+BO^2-AB^2}{2*AO*BO}\\ \\= \frac{1}{3}$

4) Since, electric field is vector quantity.

Magnitude of Net field due to both the charges is given by

$R= \sqrt{E^2+E^2+2E^2cos(\angle AOB) }\\ \\= \sqrt{2E^2(1+1/3)} =\frac{ 2\sqrt{2}E}{ \sqrt{3} }\\ \\ =\frac{8 \sqrt{2}kq}{3 \sqrt{3}x^2}$

This is our required net field whose direction is shown in figure which bisects angle AOB in the plane of triangle AOB.