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(A) The relationship between electric field $\overrightarrow{E}$ and electric potential $V$ is given by the formula: $dV = -\overrightarrow{E} \cdot d

Vedclass · Accepted Answer

$-80$

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(A) The relationship between electric field $\overrightarrow{E}$ and electric potential $V$ is given by the formula: $dV = -\overrightarrow{E} \cdot d\overrightarrow{r}$.For a one-dimensional field along the $x$-axis,this becomes: $dV = -E_x dx$.To find the potential difference $V_A - V_O$,we integrate from the origin $(x = 0)$ to point $A$ $(x = 2 \, m)$:$V_A - V_O = -\int_{0}^{2} E_x dx$.Substituting the given electric field $E_x = 30x^2$:$V_A - V_O = -\int_{0}^{2} 30x^2 dx$.Evaluating the integral:$V_A - V_O = -30 \left[ \frac{x^3}{3} ight]_{0}^{2}$.$V_A - V_O = -30 \left( \frac{2^3}{3} - \frac{0^3}{3} ight)$.$V_A - V_O = -30 \left( \frac{8}{3} ight) = -10 	imes 8 = -80 \, V$.

Assume that an electric field $\overrightarrow{E} = 30x^2 \hat{i}$ 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: (in $, V$)

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