You charge a 2.00-μF capacitor to 50.0 V. How much additional energy must you add to charge it to 100 V?

The additional energy required to be added to charge the 2μF from 50V to a 100V is 0.0075J.

Explanation:

Energy of a capacitor is given by

E = CV²/2 where C is capacitance = 2μF = 2 × 10⁻⁶ F

When the capacitor is charged to 50V, the energy it has is,

E = (2 × 10⁻⁶ × 50²)/2 = 0.0025 J

When the capacitor is charged to 100V, the energy it has stored,

E = (2 × 10⁻⁶ × 100²)/2 = 0.01 J

Energy difference between being charged to 50V and being charged to 100V is,

0.01 - 0.0025 = 0.0075 J

This represents the additional energy required to be added to charge the 2μF from 50V to a 100V.

Hope this Helps!!!

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explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

massxgravity x heght=weightxheight= pe

4900=490xh

h=10