A parallel plate capacitor has a capacitance of 6.78 μF and it is connected to a 12V battery. What is the charge on the capacitor?

8.136×10⁻⁵ J

Explanation:

Applying,

Q = Cv Equation 1

Where Q = Charge on the capacitor, v = voltage of the battery, C = capacitance of the capacitor.

From the question,

Given: C = 6.78μF = 6.78×10⁻⁶ F, v = 12 V

Substitute these values into equation 1

Q = (6.78×10⁻⁶ )(12)

Q = 8.136×10⁻⁵ J

Hence the charge on the capacitor is 8.136×10⁻⁵ J

b.

hope this : )

0.11 m/s and 0.28 m/s should be correct