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