6-3 The current through a 0.1-μF capacitor is a rectangular pulse with an amplitude of 2 mA and a duration of 5 ms. Find the capacitor voltage at the end of the pulse when the capacitor voltage at the beginning of the pulse is - 1 V.

999 V

Explanation:

Charge deposited on the capacitor

= current x time

= 2 x 10⁻³ A x 5 x 10⁻³

= 10⁻⁵ C

increase in volt = charge deposited / capacitance

10⁻⁵ /.1 x 10⁻⁶

= 10² V

= 100 V

voltage in the beginning = - 1 V

voltage at the end

= - 1 + 100

99 V.