Question

The elctric potential in a region is given by v = 3x^2 + 4x .An electron is placed at the origin will experience​

Answer 1

Explanation:

E=-(dV/dx)=-(6x+4)

at origin so x=0

E=-4

F=qE=-1.6×10^(-19)×(-4)=6.4×10^(-19) N

Answer 2

Answer:

$6.4 \times 10^{-19} \ N$

Explanation:

$v=3x^{2} +4x$

⇒ $\frac{dv}{dx} = 3(2x)+4$

⇒ $6x+4$

We know that the formula for the electric field goes as follows,

$E = -\frac{dv}{dx}$

$E = -(6x+4)$

at origin,

$x=0$

$\vec{E} = -4 \^{i} \ N/C$

$Force = Charge \times Electric \ Field$

Substituting the values in the above formula we get,

$= 1.6 \times 10^{-19} \times (-4)$

$= 6.4 \times 10^{-19} \ N$ along the positive $x$ axis.

Final Answer:

The electric potential in a region is given by $v = 3x^2 + 4x$. An electron placed at the origin will experience​ $= 6.4 \times 10^{-19} \ N$ force.

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