Assume that an electric field E = 30x^2\, i N/C exists in space. If V_0 is the potential at the origin and V_A is the potential at x = 2 m, then the potential difference V_A - V_0 is

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Quantrex Academy · Accepted Answer

Correct answer: A. -80\ V

The relation between electric potential and electric field is given by

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

.

Since the electric field is only in the x-direction, we have

dV = -E_x dx

.

Integrating both sides from the origin (x = 0

) to

x = 2

m:<br><br>\int_{V_0}^{V_A} dV = -\int_{0}^{2} 30x^2 dx

V_A - V_0 = -30 \left[ \dfrac{x^3}{3} \right]_{0}^{2}

V_A - V_0 = -10 (2^3 - 0^3)

V_A - V_0 = -80

V.

Answer:

-80\ V

Assume that an electric field E = 30x^2\, i N/C exists in space. If...

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