Verified Answer :
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:
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 \(-3.01\times 10^{-28}\ volts\).