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Electricity

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6 Oct 2008 12:25:28 IST

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easy..elec..

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electric potential is given by

V=6x-8xy^2-8y+6yz-4z^2-4x^2. then elecric force acting on point charge of 2C placed at origin will bbe

a)2N b)6N c)8N d)20N

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Priyesh

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Joined: 18 Feb 2007

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6 Oct 2008 12:50:43 IST

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V=6x-8xy^2-8y+6yz-4z^2-4x^2.

E = - (partial diff of V with x + partial diff of V with y + partial diff of V with z)

= - ( (6 - 8y^2 - 8x) + (-16xy - 8 + 6z) + ( 6y-8z))

now since particle is at origin hence x,y,z =(0,0,0)

E = -(6-8) = -(-2) = 2

so force acting =qE = 2*2 = 4N

niti agrawal

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Posts: 169

6 Oct 2008 13:10:29 IST

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sorry but no such optiion available i too got the same ans

niti agrawal

Hot goIITian

Joined: 5 Feb 2008

Posts: 169

7 Oct 2008 19:42:19 IST

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V=6x-8xy^2-8y+6yz-4z^2-4x^2

electric field = - partial derivative of V wrt x ,y &z

E=-(6-8y^2-8x)i^-(-16xy-8+6z)j^-(6y-8z)z^

at origin x=y=z=0

E=-6i^+8j^

magnitude of E=sqrt(6^2+8^2)=10

F=qE =2x10 =20 N

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