How does electric current change electric potential?

[Apteronotus]

Suppose we have an electrically insulated box in a room, and the potential difference between the box and the room is given by $$V_1=V_{Box1}-V_{room}$$.

Now suppose we inject the box with X amps of current, for t seconds. The new voltage is now given by $$V_2=V_{Box2}-V_{room}$$

By how much does the injection of X amps of current change the electric potential of the box?
ie.
What is $$V_{Box1}-V_{Box2}$$?

 

[Stonebridge]

It depends on the capacitance of the box. V=Q/C
and Q = It
Your current has placed more charge on the box so its potential will rise according to the value of its capacitance.
Edit
Use V=Q/C and write an expression for V1 and V2 in terms of the charge on the box and its capacitance.

 

[Apteronotus]

So can we equate the two equations you've written and simply say $$CV=It$$?

Also,
if the box is a good electrical insulator, then we expect it to have a high capacitance (ability to hold an electrical charge), right?
But then by the CV=It above, the higher the C is the less effect a current I will have on the potential V. What's the intuitive meaning behind this?

 

[Stonebridge]

So can we equate the two equations you've written and simply say $$CV=It$$?

The initial pd on the box is given by V1=Q/C where Q is the initial charge and C its capacitance.
The final pd of the box (V2) is found by putting the charge on it equal to Q + It (and the same value of C.)
You then find V2 - V1

Also,
if the box is a good electrical insulator, then we expect it to have a high capacitance (ability to hold an electrical charge), right?
But then by the CV=It above, the higher the C is the less effect a current I will have on the potential V. What's the intuitive meaning behind this?

The definition of C is that it equals Q/V
The intuitive part of this is that the higher the value of C, the more charge you need to increase the pd. So as current is flow of charge, the higher the capacitance means that you need more current to increase the pd.
I always think of water in a container. The pd is the depth of the water, and the capacitance is the capacity (volume) of the container. The bigger the capacity of the container, the more water you need to raise its level.
If you were filling it with a hose pipe, you would need a larger water current to fill a larger container to the same depth in the same time.

 

[sardar]

Electric current is the flow of electric charge through a conductor. When a current flows through a material, it creates an electric potential difference, or voltage, between different points in the material. This voltage is a measure of the energy required to move a unit of charge from one point to another.

In the scenario described, the electric potential difference between the box and the room is initially V_1. When X amps of current is injected into the box for t seconds, the new voltage becomes V_2. This change in voltage is due to the movement of electric charge within the box.

The exact amount by which the injection of X amps of current changes the electric potential of the box can be calculated using Ohm's law, which states that the voltage (V) is equal to the current (I) multiplied by the resistance (R) of the material: V = I * R. In this case, the resistance of the box remains constant, so we can rearrange the equation to solve for the change in voltage (ΔV): ΔV = I * R * t.

Therefore, the injection of X amps of current for t seconds will result in a change in voltage of ΔV = X * R * t. This means that the electric potential difference between the box and the room will change by this amount, or V_{Box1}-V_{Box2} = X * R * t.

In summary, the injection of electric current into the box will change the electric potential of the box by an amount determined by the current, resistance of the material, and the time for which the current flows.