Question

Given a cube with point charges q on each of its vertices.calculate the force exerted on any of the charges due to the rest of the seven charges.


Answer:-kq^2/a^2 [(1/3root3 +1/root 2 +1) (icap+j cap+k cap]
please someone solve this question

Answer 1

Given:

Given a cube with point charges q on each of its vertices.

To find:

Calculate the force exerted on any of the charges due to the rest of the seven charges.

Solution:

From given, we have,

A cube with point charges q on each of its vertices.

The force exerted by charges is given as follows:

F12 = k q1q2/|r1 - r2|³ (r1 - r2)

So, we have,

Fa1 = kq²/a³ (-ak)

Fa2 = kq²/2√2a³ (-ak - aj)

Fa3 = kq²/3√3a³ (-ai - ak - aj)

Fa4 = kq²/2√2a³ (-ai - ak)

Fa5 = kq²/a³ (-ai)

Fa6 = kq²/2√2a³ (-ai - aj)

Fa7 = kq²/a³ (- aj)

Therefore, the resultant force is given by,

Fres = Fa1 + Fa2 + Fa3 + Fa4 + Fa5 + Fa6 + Fa7

∴ Fres = - kq²/a³ [ 1/3√3 + 1/√2 + 1 ] (i + j + k)