# Five charge q each are placed at the corners of regular pentagon os side a what will be the electric field at O if the charge q at A is replaced by -q ? solve it

1
by hkri5parublah

2016-05-03T17:48:24+05:30

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See diagram ...

E1 = E2 = E3 = E4 = E5 in magnitude.

E1 = (1/4πε) * q/ R²
For the pentagon,   sin 36° = (a/2)/R
R = a/(2 sin 36°) = 0.85 a  approx.

E1 = (1/4πε) * (4 q sin² 36°) / a²

The resultant of E1 and E5 is along OB from symmetry and is
= E6 = 2 * E1 Cos 36°

Similarly, the resultant of E4 and E3 is along OE and is
E7 = 2 * E1 * Cos 36°

Resultant of E6 and E7 is along OA from symmetry and is =
E8 = 2 * (2 * E1 * Cos 36°) * Cos 72°
= 4  * E1 * Cos 36° * Cos 72°

Now we can add E1 due to charge at A and E8.
Resultant =  E1 * [ 1 + 4 Cos 36 Cos 72 ]

=  (1/4πε) * [ (4 q Sin² 36) / a² ] * [1 + 4 Cos 36 cos 72]

along the direction OA.

you can simplify the answer ..if needed.

=  (1/4πε) * 2.764  * q/ a²   approximately..