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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