Question

A proton is moved so that its electric potential energy increases from 2.0 × 10-18 J to 6.0 × 10-18 J. The magnitude of the charge on a proton is 1.602 × 10-19 C.

The electric potential difference through which the proton moved, to the nearest whole number, is _______V

Answer 1

Answer : The electric potential difference through which the proton moved, to the nearest whole number is, 25 V

Solution :

Potential difference : It is defined as the work done by the unit charge.

Formula used :

$\Delta V=\frac{\Delta E}{q}$

where,

$\Delta V$ = electric potential difference

$\Delta E$ = electric potential energy

q = charge on  proton

Now put all the given values in the above formula, we get the value of electric potential difference.

$\Delta V=\frac{\Delta E}{q}=\frac{(6\times 10^{-18}-2\times 10^{-18})J}{1.602\times 10^{-19}C}=24.968V=25V$

Therefore, the electric potential difference through which the proton moved, to the nearest whole number is, 25 V

Answer 2

The potential difference is  25 V through which the proton moved. The electrical difference between two points is called Potential difference.

What is the Potential difference?

It is defined as the electrical difference between two points. It can be calculated by the formula,

$\Delta V = \dfrac{ \Delta E} {q}$

Where,

$\Delta V$ = electric potential difference

${ \Delta E}$ = electric potential energy = $6.0 \times 10^{-18} { \rm \ J} - 2.0 \times 10^{-18} { \rm \ J} = \bold{4.0 \times 10^{-18} { \rm \ J}}$

$q$ = charge on  proton = $1.602 \times 10^{-19} { \rm \ C}$

Put the values in the formula,

$\Delta V = \dfrac{{4.0 \times 10^{-18} { \rm \ J}} } {1.602 \times 10^{-19} { \rm \ C}}\\\\\Delta V = 25 \rm \ V$

Therefore, the potential difference is  25 V through which the proton moved.

Learn more about the potential difference:

https://brainly.com/question/1313684