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A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is:

  • Option 1)

    {{\text{E}^2 \text{Ad}} \mathord{\left/ {\vphantom {{\text{E}^2 \text{Ad}} {\varepsilon _\text{O} }}} \right. \kern-\nulldelimiterspace} {\varepsilon _\text{O} }}

  • Option 2)

    \frac{1} {2}\varepsilon _\text{O} \text{E}^2 \text{Ad}

  • Option 3)

    \varepsilon _\text{O} \text{E}^2 \text{Ad}

  • Option 4)

    \frac{1} {2}\varepsilon _\text{O} \text{E}^2

 

Answers (1)

best_answer

As we learnt in 

Parallel Plate Capacitor -

C=frac{epsilon _{0}A}{d}

- wherein

Area - A seperation between two plates.

 

 

Potential difference the between plates

V=Ed

u = \frac{1}{2}cv^{2} =\frac{1}{2}\left ( \frac{\varepsilon _{0}A}{d} \right )\left ( Ed \right )^{2}

u= \frac{1}{2}\varepsilon _{0}E^{2}A.d


Option 1)

{{\text{E}^2 \text{Ad}} \mathord{\left/ {\vphantom {{\text{E}^2 \text{Ad}} {\varepsilon _\text{O} }}} \right. \kern-\nulldelimiterspace} {\varepsilon _\text{O} }}

Incorrect

Option 2)

\frac{1} {2}\varepsilon _\text{O} \text{E}^2 \text{Ad}

Correct

Option 3)

\varepsilon _\text{O} \text{E}^2 \text{Ad}

Incorrect

Option 4)

\frac{1} {2}\varepsilon _\text{O} \text{E}^2

Incorrect

Posted by

Aadil

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