A parallel-plate capacitor has dimensions 4.3 cm × 4.6 cm. The plates are separated by a 1.0-mm thickness of paper (dielectric constant κ = 3.7). What is the charge that can be stored on this capacitor, when connected to a 1.2-V battery? (ε0 = 8.85 × 10−12 C2/N⋅ m2)

7.77×10^-11coulombs

Explanation:

Firstly, we will calculate the capacitance of the capacitor (C) which is given mathematically as;

C =κε0A/d where;

κ is the dielectric constant= 3.

ε0 is the permittivity of free space = 8.85 × 10^−12 C²/Nm²

d is the distance between the plates = 1.00mm = 0.001m

A is the area = 4.3cm×4.6cm = 0.043m×0.046m

A = 0.001978m²

C=3.7× 8.85×10^-12×0.001978/0.001

C = 6.48×10^-11Farads

The charge on the capacitor when connected to 1.2V battery is gotten by using the formula;

Q = CV where

C is the capacitance of the capacitor

V is the voltage across the capacitor

Q = 6.48×10^-11×1.2

Q = 7.77×10^-11Coulombs

are there options? cause you ain’t. real specific

0.19 m/s and 0.45 m/s