A) \[2\times {{10}^{-6}}N\]
B) \[4\times {{10}^{-6}}N\]
C) \[6\times {{10}^{-6}}N\]
D) \[8\times {{10}^{-6}}N\]
Correct Answer: D
Solution :
Given, \[V=5+4{{x}^{2}}\] \[\frac{dV}{dr}=E\] \[\therefore \] \[\frac{dV}{dx}=8x=E\] or \[E=8\times 0.5=4\,V{{m}^{-1}}\] Now, \[F=qE\] \[=2\times {{10}^{-6}}\times 4\] \[=8\times {{10}^{-6}}N\]
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