Q.
Given Electric potential at a point is: V=x2y+yz . The electric field at the point (1,3,1) is
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a
7 units
b
70 units
c
49 units
d
490 units
answer is A.
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Detailed Solution
electric field isE= - [dVdxi^ +dVdyj^ + dvdzk^]E = -2xyi^ + (x2+ z)j^+ yk^put x=1, y=3, z=1 in above equation E→=-[2(1)(3)i^+(12+1)j^+3k^=-[6i^+(2)j^+3k^]E=62+22+32=7 units
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