Question

Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is charged and stores 8.49 nJ of energy. Find the electric field strength inside the capacitor.

Answer 1

Answer:

E= 55.53 x 10³ V/m

Explanation:

Given that

a=  3.63 cm

Area ,A= a²

distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Value of capacitor given as

$C=\dfrac{\varepsilon _oA}{d}$

By putting the values

$C=\dfrac{8.85\times 10^{-12}\times 3.63^2\times 10^{-4}}{0.473\times 10^{-3}}$

C=2.46 x 10⁻¹¹ F

$U=\dfrac{1}{2}CV^2$

V=Voltage difference

$V=\sqrt{\dfrac{2U}{C}}$

$V=\sqrt{\dfrac{2\times 8.49\times 10^{-9}}{2.46\times 10^{-11}}}$

V=26.27 V

V= E d

E=Electric filed

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

Answer 2

The electric field strength inside the capacitor is E= 55.53 x 10³ V/m

Calculation of electric field;

Since a=  3.63 cm

Area ,A= a²

distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Now

Capacitor value should be $C = \varepsilon_oA \div d$

Now

$C = \frac{8.85\times 10^{-12}\times 3.63\times 10^{-4}}{0.473\times 10^{-3}}$

Now

$U = 1 \div 2 CV^2\\\\V = \sqrt{2U \div C}\\\\ = \sqrt{2\times 8.49\times 10^{-9} \div 2.46 \times 10^{-11}}$

= 26.27V

Now

The electric field should be

V= E d

So,

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

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