An electric field (in $V / m$ ) is given by $E = 10 x^{3}$ . Determine the potential difference, in volts, between $x = 0 m$ and $x = 3 m$ .

202.5

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Solution

The correct option is B $202.5$
We know that, $E = - \frac{d V}{d x}$

$V = - \int_{0}^{3} E d x = - \int_{0}^{3} 10 x^{3} d x = - 10 \times \frac{3^{4}}{4} = 202.5$

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An electric field (in V/m) is given by E=10x3. Determine the potential difference, in volts, between x=0m and x=3m.

The correct option is B 202.5We know that, E=−dVdxV=−∫30Edx=−∫3010x3dx=−10×344=202.5

An electric field (in $V / m$ ) is given by $E = 10 x^{3}$ . Determine the potential difference, in volts, between $x = 0 m$ and $x = 3 m$ .

The correct option is B $202.5$
We know that, $E = - \frac{d V}{d x}$

$V = - \int_{0}^{3} E d x = - \int_{0}^{3} 10 x^{3} d x = - 10 \times \frac{3^{4}}{4} = 202.5$