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Hint: The potential energy of a charge depends on the sign of the charge as well as the direction of the electric field which is exerting on the charged particle. It will be the combination of both these events which decides whether the electrical potential energy of the charge will increase or decrease.
Formula used: In this question, we will use the following formula:
The potential energy of a charge \[U = q\Delta V\] where $q$ is the charge and $\Delta V$ is the change in potential experienced by the charge.
Complete step by step answer:
We know that when a positive charge moves from a region of higher to lower potential, it will lose its potential energy. This is because the tendency of the positive charge to move in the system decreases with decreasing potential. Mathematically, this can be explained from the formula of potential energy as
\[U = q\Delta V\]
Now, since the charge moves from a region of high to low potential the term $\Delta V$ will be positive. Since in this case, the charge is also positive, so will the work be done.
However, if the charge is negative, the change in potential energy will be negative since $q$ will be negative.
Hence the potential energy may increase or decrease depending on the sign of the charge.
So, option (C) is correct.
Note: If the potential energy of the system is positive, in this case, for a positive charge, the electric potential or the corresponding electric field will be responsible for doing work on the charge and moving it. If the charge was negative, the potential energy would also be negative and we would have to do work against the system to move the charge.
Formula used: In this question, we will use the following formula:
The potential energy of a charge \[U = q\Delta V\] where $q$ is the charge and $\Delta V$ is the change in potential experienced by the charge.
Complete step by step answer:
We know that when a positive charge moves from a region of higher to lower potential, it will lose its potential energy. This is because the tendency of the positive charge to move in the system decreases with decreasing potential. Mathematically, this can be explained from the formula of potential energy as
\[U = q\Delta V\]
Now, since the charge moves from a region of high to low potential the term $\Delta V$ will be positive. Since in this case, the charge is also positive, so will the work be done.
However, if the charge is negative, the change in potential energy will be negative since $q$ will be negative.
Hence the potential energy may increase or decrease depending on the sign of the charge.
So, option (C) is correct.
Note: If the potential energy of the system is positive, in this case, for a positive charge, the electric potential or the corresponding electric field will be responsible for doing work on the charge and moving it. If the charge was negative, the potential energy would also be negative and we would have to do work against the system to move the charge.
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