Question

A proton is moved from a position where the electric potential is 125 V to a position where the electric potential is 275 V. The magnitude of the charge on a proton is 1.602 × 10-19

c.

Answer 1

q = 2.403 x 10-7 F

Answer 2

We are asked to solve for the capacitance of a charged proton and the formula is shown below:
C= q / ΔV

The given values are the following:
ΔV = 275 volts - 125 volts
C= 1.602 x 10-9 C
q = C * ΔV
q = 1.602x10-9 C * (150 volts)
q = 2.403 x 10-7 F