The electric potential V is given as a function of distance x by V=5x2+10x−9. The value of electric field at x=1 is (All quantities are in SI units.)
−20
6
11
−23
A
−20
B
−23
C
11
D
6
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Solution
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V=5x2+10x−9
→E=−∂V∂x^i−∂V∂y^j−∂V∂z^k
Now, ∂V∂x=10x+10
and ∂V∂y=∂V∂z=0
Hence, →E=−(10x+10)^i
For x=1, →E=−20^i
Answer-(A)
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