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.

Question

Physics

Rex Marshall

3 January, 03:25

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.

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Ryker Jarvis · Answer 1

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

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 * 10-19c.

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.