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
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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..
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..
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Answer:
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Explanation:
symmetical figures the Electric feild at center is zero
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