Respuesta :
Answer:
-600 V
Explanation:
The expression for the relationship between work done and a charge is as follows;
[tex]W=q\Delta V[/tex]
Here, W is the work done, q is the charge and [tex]\Delta V[/tex] is the change in potential.
It is given in the problem that a charged particle is released from rest at point A.
The change in kinetic energy is equal to the work done. The kinetic energy is zero as the velocity of a charged particle is zero at point A.
The relation between change in kinetic energy and work done is as follows;
[tex]W= KE_{f}-KE_{i}[/tex]
Here, [tex]KE_{f}[/tex] is the final kinetic energy and [tex]KE_{i}[/tex] is the initial kinetic energy.
According to question,
[tex]W= KE_{f}[/tex]
W= 4.8 J
Use the equation (1) to get the value of electric potential difference.
[tex]W=q\Delta V[/tex]
Put [tex]\Delta V= V_{B}-V_{A}[/tex], W= 4.8 J and [tex]q = –8.0 mC=8\times 10^{-3}C[/tex].
[tex]4.8=V_{B}-V_{A}(8\times 10^{-6}) [/tex]
[tex]V_{B}-V_{A}= \frac{4.8}{-8\times 10^{-3}}[/tex]
[tex]V_{B}-V_{A}= {-6\times 10^{-7}}V[/tex]
[tex]V_{B}-V_{A}= -600V[/tex]
Therefore, the electric potential is -600 V.
