Energy density (in vacuum) at a place where electric field intensity $E$ is:

\frac{1}{2} ε_{0} E

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\frac{E^{2}}{2 ε_{0}}

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\frac{1}{2} E ε_{0}^{2}

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\frac{1}{2} ε_{0} E^{2}

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Solution

The correct option is A $\frac{1}{2} ε_{0} E^{2}$
Energy density $(u_{d})$
$u_{d} = \frac{1}{2} ε_{0} E^{2}$

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Similar questions

E

If e is the electric field intensity of an electrostatic field then the electrostatic energy density is proportional to

1)e

2)e×e

3)1/e×e

4)e×e×e

Answer is 2

(\frac{1}{2}) ε_{0} E^{2}

ε_{0}

(\frac{1}{2}) ε_{0} E^{2}

ε_{0}

ε_{0}

E

\frac{1}{2} ε_{0} E^{2}

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