Respuesta :
Answer:
Potential difference between A and B is [tex]\Delta V = -3.01\times 10^{-28}\ volts[/tex]
Explanation:
It is given that,
Charge, [tex]q=7.3\times 10^{15}\ C[/tex]
The charge moves from point A to B is an electric field.
Potential energy at A, [tex]V_A=3.5\times 10^{-12}\ J[/tex]
Potential energy at B, [tex]V_B=1.3\times 10^{-12}\ J[/tex]
We have to find potential difference between A and B. The relationship between potential energy and potential difference is given by :
[tex]\Delta U=q\Delta V[/tex]
Where
[tex]\Delta U[/tex] is change in potential energy between A and B
[tex]\Delta V[/tex] is change in potential difference between A and B
[tex]\Delta V =\dfrac{\Delta U}{q}[/tex]
[tex]\Delta V=\dfrac{1.3\times 10^{-12}\ J-3.5\times 10^{-12}\ J}{7.3\times 10^{15}\ C}[/tex]
[tex]\Delta V = -3.01\times 10^{-28}\ volts[/tex]
Hence, this is the required solution.
Answer:
The potential difference between point A and B is [tex]\Delta V=-3.01\times 10^{-28}\ volts[/tex].
Explanation:
Given that,
Charge, [tex]q=7.3\times 10^{15}\ C[/tex]
Potential energy at point A, [tex]U_A=3.5\times 10^{-12}\ J[/tex]
Potential energy at point B, [tex]U_B=1.3\times 10^{-12}\ J[/tex]
To find,
The potential difference between point A and B or [tex]\Delta V[/tex]
Solve,
We know that the relationship between the potential difference and the potential energy is given by :
[tex]\Delta U=q\times \Delta V[/tex]
Where
[tex]\Delta U[/tex] is the difference in potential energy between A and B
[tex]\Delta V[/tex] is the potential difference between point A and B
[tex]\Delta V=\dfrac{U_B-U_A}{q}[/tex]
[tex]\Delta V=\dfrac{1.3\times 10^{-12}-3.5\times 10^{-12}}{7.3\times 10^{15}}[/tex]
[tex]\Delta V=-3.01\times 10^{-28}\ volts[/tex]
Therefore, the potential difference between point A and B is [tex]-3.01\times 10^{-28}\ volts[/tex].
