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I was learning about electric potential and it was pointed out by my professor that the potential we calculate between two points is the potential difference and not the absolute potential; if potential is defined as the work done per unit charge or test charge, how does the work done to move it from point b to a (assuming that it took some work to bring the charge to b, against the electric field) give the potential difference and not the potential?

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6 Answers 6

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I was learning about electric potential and it was pointed out by my professor that the potential we calculate between two points is the potential difference and not the absolute potential

It's correct. Absolute potential is defined as the amount of work that needs to be done to bring a positive charge to its current location from a point of infinite distance. On the other hand potential difference is defined as if a positive charge is taken from a point to another point in an electric field then the difference of the electric potential between the two points is called the potential difference.

Let's understand it with an example. Suppose there are two points $a$ and $b$ in an electric field. A positive charge is taken from point $a$ to point $b$ against the electric field. Then we can say that some amount of work is done. Let's assume the work is $W$ and the charge is $q$. Then the electric potential, $V = \frac {W}{q}$

Suppose the electric potential of point $a$ was $V_a$ and the electric potential of $b$ is $V_b$ then the electric potential difference, $V_{ba} = V_b - V_a = V = \frac {W}{q}$ (Note that $b$ has higher electric potential than $a$)

Hope this helps.

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Answer is simple. Because we measure it from infinity, it's also the potential difference I agree, but we take an assumption there that the potential at the infinity might be zero.

So while calculating for $V_r-V_∞$ we put $V_∞=0$ so that's how we only get the value for $V_r$. However, it's not necessary to have zero potential at the infinity, it can be anything. The thing that matters the most is potential difference not the potential.

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Assume a test charge (unit positive charge) is placed at a distance of $r_b$ from a point charge having charge $+q$
Now this test charge is moved so that now it is at $r_a$ from the charge $q$

Force on test charge due to $q$ : $$\vec{F}=q_{\text{t}} \vec{E} = k\frac{q}{r^2}$$ $q_t = \space \text{Charge on test charge = 1C}$
$r=\text{Distance of test charge from q}$

Since Electrostatic force is conservative, it has some potential energy associated with it.

$$W_{\text{Conservative Foce}}=-\Delta U$$ $$\vec{F}.\vec{dr}=-dU$$ $$k\frac{q}{r^2}dr=-dU$$ $$kq\int_b^a \frac{1}{r^2}dr=-\int_b^adU$$ $$kq\frac{1}{r_b} - kq\frac{1}{r_a}=U_b-U_a$$

$k\frac{q}{r_a}$ is potential (or work done in bringing test charge to that point) of position $a$ when charge is brought from infinity to $a$. Similarly for position $b$.

So, $$\Delta U_{\text{b to a}} = V_a-V_b$$ $$-W_{\text{int. cons. force}}=W_{\text{ext}}=\Delta U = V_a-V_b$$

Above case applies when the test charge is brought very slowly (using external force) so that only potential energy changes and Kinetic energy remains $0$. In such case external force is equal to electrostatic force (internal conservative force) at all points of path.
Thus, $-W_\text{electrostatic force} = W_{\text{external}}$

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  • $\begingroup$ I have corrected the errors in my answer ! Please reconsider your vote. $\endgroup$
    – Spencer
    May 31 at 10:11
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. . . . if potential is defined as the work done per unit charge or test charge . . . .

It is not as this definition is incomplete and is the reason for you asking your question.

It can be defined as,

the potential at a point is the work done in taking unit positive charge from a (defined) position of zero potential to the point.

The defined position of zero potential does not have to be infinity as it could be the negative terminal of a voltage source, a plate of a capacitor, the surface of the Earth, . . . . .

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Work done against the field in moving a unit charge from a to b : $$\int_{a}^{b} -\vec{E} \cdot \vec{dr}$$

From the fundamental theorem of line integrals

$$V(b)-V(a) = \int_{a}^{b} \nabla V \cdot \vec{dr}$$

Because $-\vec{E}$ Is conservative, we know that this can be written as the gradient of some scalar function, $$-\vec{E} = \nabla V$$

Substituting this into the fundamental theorem of line integrals

$$V(b)-V(a) = \int_{a}^{b} -\vec{E} \cdot \vec{dr}$$

Thus the potential difference of a and b is the work done per against the field in moving from a to b. This qauntity is independant of the path inbetween as the E field is Conservative

Now it is just a matter of finding the correct V to use, which can be done by the definition of E and V

Edit:

$$V(b)-V(a) = \int_{a}^{b} -\vec{E} \cdot \vec{dr}$$

The right hand side can be split up into 2 integrals

$V(b)-V(a)$= $$(\int_{ref}^{b} -\vec{E} \cdot \vec{dr})-( \int_{ref}^{a} -\vec{E} \cdot \vec{dr})$$

Thus it follows that:

$$V(r) = \int_{ref}^{r} -\vec{E} \cdot \vec{dr}$$

Here it can be shown that potential is non uniquely defined, there can be many references of potential, as the difference between 2 potential functions is invariant. This is also indicative of the +c when directly solving for the potential. In most situations, the zero of potential is infinity, or the negative plate of a battery

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Electric potential is defined as potential energy per unit charge(i.e how much work is done to move a unit positive charge from infinity to the particular point in electric field).

Now , imagine there are two points $a$ and $b$ . These two points are seperate from each other , so obviously there must be the difference between the electric potential of these two points.

Let point $a$ be the point of high potential and point $b$ be the point of low potential. If a charge is moved from $a$ to $b$ , there must be change in its potential energy (because electric potential is the potential energy of unit positive charge).

We know that change in energy is equal to work done ($∆E= W.D$)

electric potential is also the potential energy of unit positive charge , so change in electric potential must be equal to work done on unit positive charge.

Hope this helps

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    $\begingroup$ Actually change in energy is equal to work done. The equation should be $∆E = F_c \cdot d$ (where $F_c$ is the force by which work is done) $\endgroup$ May 31 at 7:09
  • $\begingroup$ @DebanjanBiswas yes, the work must be done by some external agent that's why it's called potential difference, because the workdone by the charge alone has no use. And yes negative of work done by a conservative force is what we call potential difference. Both the things holds true. $\endgroup$ May 31 at 8:34

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