C,O.(A)(B)Figure 4:4. The energy stored in a capacitor is U = = }CV² = }Qv. The energy per unit volume for a vacuum filled capacitor is up = tco E².If the electric field is constant within the capacitor of volume I the energy stored follows as U = u pT. If the electric field is not constant thenU = | drup. It follows that for a parallel plate capacitor with area A = 100cm2 and plate separation h = lem connected to a battery of voltageV = 5000V olts the energy stored is[ignore the fringing] U = ugT = }60E²+ = }co¥ Ah = }co A = 0.111mJ and therefore the magnitudeof the force on one plate is F = = }o A = 11.1mN (True,False)

The energy stored be calculated as, U=12ε0V2hA=128.854×10-12 F.m-15000 V21 cm100 cm2=128.854×10-12…

4. The energy stored in a capacitor is U = = }cv? = }Qv. The energy per unit volume for a vacuum filled capacitor is u gE = c0E?. %3D If the electric field is constant within the capacitor of volume r the energy stored follows as U = ugT. If the electric field is not constant then U = [ drug. It follows that for a parallel plate capacitor with area A = 100cm? and plate separation h = lem connected to a battery of voltage V = 5000V olts the energy stored islignore the fringing] U = ugT = tcoE?7 = }eo Ah = to A = 0.111mJ and therefore the magnitude of the force on one plate is F = = oA = 11.1mN (True,False) %3D

4. The energy stored in a capacitor is U = = }cv? = }Qv. The energy per unit volume for a vacuum filled capacitor is u gE = c0E?. %3D If the electric field is constant within the capacitor of volume r the energy stored follows as U = ugT. If the electric field is not constant then U = [ drug. It follows that for a parallel plate capacitor with area A = 100cm? and plate separation h = lem connected to a battery of voltage V = 5000V olts the energy stored islignore the fringing] U = ugT = tcoE?7 = }eo Ah = to A = 0.111mJ and therefore the magnitude of the force on one plate is F = = oA = 11.1mN (True,False) %3D