Assume that the electric field E = 30 x^2 i exists in space. Then the potential difference V_A - V_O , where ' V_O ' is the potential at the origin and ' V_A ' is the potential at x = 2 m, is

Question

Quantrex Academy · Accepted Answer

Correct answer: D. -80 V

The relation between electric potential and electric field is given by

dV = -\vec{E} \cdot d\vec{r}

.

Substituting the given electric field

\vec{E} = 30x^2 \hat{i}

and

d\vec{r} = dx \hat{i}

:

dV = -(30x^2 \hat{i}) \cdot (dx \hat{i}) = -30x^2 dx

Integrating both sides from the origin (x = 0

) to point A (x = 2

):

\int_{V_O}^{V_A} dV = \int_{0}^{2} -30x^2 dx

V_A - V_O = -30 \left[ \dfrac{x^3}{3} \right]_{0}^{2}

V_A - V_O = -10 (2^3 - 0^3)

V_A - V_O = -10 \times 8 = -80

V

Answer:

-80

V

Assume that the electric field E = 30 x^2 i exists in space. Then...

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