Question

♡HLO✌


Four particles, each having a charge q, are placed on the four vertices of a regular pentagon. The distance of each corner from the centre is "a ". Find the electric field at the centre of the pentagon. ✔✔


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Answer 1

Assume:

There were 5 charges each on the corner.

In that case:

The electric field at the centre would be zero. On removing one of the charges the resultant of the remaining 4 vectors is equal to the magnitude of the removed vector.

And:

The direction of net field is along -E(E).

Note: Check these attachments.

Therefore:

$\boxed{\sf{E(E) = \frac{1}{4\pi\in_{0}} \frac{q}{ {a}^{2} }}}$

Answer 2

Answer:

Assume:

There were 5 charges each on the corner.

In that case:

The electric field at the centre would be zero. On removing one of the charges the resultant of the remaining 4 vectors is equal to the magnitude of the removed vector.

And:

The direction of net field is along -E(E).

Note: Check these