Question

A charge q of 7.3 × 1015 coulomb moves from point A to point B in an electric field. If the potential energy of the charge at point A is 3.5 × 10-12 joules and that at point B is 1.3 × 10-12 joules, what is the potential difference between point A and B?

Answer 1

Answer:

Potential difference between A and B is $\Delta V = -3.01\times 10^{-28}\ volts$

Explanation:

It is given that,

Charge, $q=7.3\times 10^{15}\ C$

The charge moves from point A to B is an electric field.

Potential energy at A, $V_A=3.5\times 10^{-12}\ J$    

Potential energy at B, $V_B=1.3\times 10^{-12}\ J$    

We have to find potential difference between A and B. The relationship between potential energy and potential difference is given by :

$\Delta U=q\Delta V$

Where

$\Delta U$ is change in potential energy between A and B

$\Delta V$ is change in potential difference between A and B

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

$\Delta V=\dfrac{1.3\times 10^{-12}\ J-3.5\times 10^{-12}\ J}{7.3\times 10^{15}\ C}$    

$\Delta V = -3.01\times 10^{-28}\ volts$

Hence, this is the required solution.      

Answer 2

Answer:

The potential difference between point A and B is $\Delta V=-3.01\times 10^{-28}\ volts$.

Explanation:

Given that,

Charge, $q=7.3\times 10^{15}\ C$

Potential energy at point A, $U_A=3.5\times 10^{-12}\ J$

Potential energy at point B, $U_B=1.3\times 10^{-12}\ J$

To find,

The potential difference between point A and B or $\Delta V$

Solve,

We know that the relationship between the potential difference and the potential energy is given by :

$\Delta U=q\times \Delta V$

Where

$\Delta U$ is the difference in potential energy between A and B

$\Delta V$ is the potential difference between point A and B

$\Delta V=\dfrac{U_B-U_A}{q}$

$\Delta V=\dfrac{1.3\times 10^{-12}-3.5\times 10^{-12}}{7.3\times 10^{15}}$

$\Delta V=-3.01\times 10^{-28}\ volts$

Therefore, the potential difference between point A and B is $-3.01\times 10^{-28}\ volts$.