Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick,

Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick, with a dielectric constant of K=3.60. The resultant electric field in the dielectric is $1.20×{10}^{6}$ volts per meter.
Compute the magnitude of the charge per unit area $\sigma$ on the conducting plate.
$\sigma =\frac{c}{{m}^{2}}$
Compute the magnitude of the charge per unit area ${\sigma }_{1}$ on the surfaces of the dielectric.
${\sigma }_{1}=\frac{c}{{m}^{2}}$
Find the total electric-field energy U stored in the capacitor.
u=J
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Malena
a) the magnitude of the charge per unit area $\sigma$ on the conducting plate:
E effective=$\frac{\sigma }{k{ϵ}_{0}}$
$\sigma =3.6\cdot 8.854{e}^{-12}\cdot 1.20{e}^{6}=3.82\cdot {e}^{-5}\frac{C}{{m}^{2}}$
b) The magnitude of the charge per unit area σ1 on the surfaces of the dielectric.
$\sigma$ (on the dielectric) $=\sigma \left(\left(\frac{1}{k}\right)-1\right)$
Not exactly what you’re looking for?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee