When a positive charge moves in the opposite direction of the electric field it is in its electric potential energy?

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.

Electric Potential Energy (U) and Electric Potential (V): (Notes from C. Erkal’s lectures PHYS 221)

Consider a parallel plate capacitor that produces a uniform electric field between its large plates. This is accomplished by connecting each plate to one of the terminals of a power supply (such as a battery).

Figure 1: An electric field is set up by the charged plates separated by a distance l. The charges on the plates are +Q and –Q.

Figure 2: An electric charge q is moved from point A towards point B with an external force T against the electric force qE.

Figure 3, 4: When it is moved through a distance d, its potential energy at the point B is qEd relative to the point A.

Figure 5: When released from B (T = 0), it will accelerate toward the lower plate. As it is moving toward the lower plate, its potential energy decreases and its Kinetic energy increases. When it reaches the lower plate (where we can choose the Potential energy to be zero), its potential energy at A is completely converted to Kinetic Energy at point B:

Note that qEd is the work done by the field as the charge moves under the force qE from B to A. Here m is the mass of the charge q, and v is its velocity as it reaches point A. Here we assumed that electric field is uniform! Work done by E field:

Let’s remember Kinetic Energy-Work theorem (Work Energy principle):

where we introduced the concept of potential energy and conservative force ( a force under which one can define a potential energy so that the work done only depends the differences of the potential energy function evaluated at the end points).

A rule of thumb for deciding whether or not EPE is increasing:

If a charge is moving in the direction that it would normally move, its electric potential energy is decreasing. If a charge is moved in a direction opposite to that of it would normally move, its electric potential energy is increasing. This situation is similar to that of constant gravitational field (g = 9,8 m/s2). When you lift up an object, you are increasing its gravitational potential energy. Likewise, as you are lowering an object, its gravitational energy is decreasing.

A General Formula for Potential Difference:

The work done by an E field as it act on a charge q to move it from point A to point B is defined as Electric Potential Difference between points A and B:

Clearly, the potential function V can be assigned to each point in the space surrounding a charge distribution (such as parallel plates). The above formula provides a simple recipe to calculate work done in moving a charge between two points where we know the value of the potential difference. The above statements and the formula are valid regardless of the path through which the charge is moved. A particular interest is the potential of a point-like charge Q. It can be found by simply performing the integration through a simple path (such as a straight line) from a point A whose distance from Q is r to infinity. Path is chosen along a radial line so that

becomes simply Edr. Since the electric field of Q is kQ/r2,

This process defines the electric potential of a point-like charge. Note that potential function is a scalar quantity as oppose to electric field being a vector quantity. Now, we can define the electric potential energy of a system of charges or charge distributions. Suppose we compute the work done against electric forces in moving a charge q from infinity to a point a distance r from the charge Q. The work is given by:

Note that if q is negative, its sigh should be used in the equation! Therefore, a system consisting of a negative and a positive point-like charge has a negative potential energy.

A negative potential energy means that work must be done against the electric field in moving the charges apart!

Now consider a more general case, which deals with the potential in the neighborhood of a number of charges as depicted in the picture below:

Let r1,r2,r3 be the distances of the charges to a field point A, and r12, r13, r23 represent the distance between the charges. The electric potential at point A is:

Example:

If we bring a charge Q from infinity and place it at point A the work done would be:

The total Electric Potential Energy of this system of charges namely, the work needed to bring them to their current positions can be calculated as follows: first bring q1 (zero work since there is no charge around yet), then in the field of q1 bring q2, then in the fields of q1 and q2 bring q3. Add all of the work needed to compute the total work. The result would be:

Finding Electric Field from Electric Potential:

The component of E in any direction is the negative of the rate of change of the potential with distance in that direction:

The symbol Ñ is called Gradient. Electric field is the gradient of electric potential. Electric field lines are always perpendicular to the equipotential surfaces.

Equipotentail Surfaces:

These are imaginary surfaces surrounding a charge distribution. In particular, if the charge distribution is spherical (point charge, or uniformly charged sphere), the surfaces are spherical, concentric with the center of the charge distribution. Electric field lines are always perpendicular to the equipotential surfaces. The equation

implies that due to the negative sign, the direction of E is opposite to the direction in which V increases; E is directed from higher to lower levels of V (from higher potential to lower potential). Another words, the gradient of a scalar (in this case E field) is normal to a surface of constant value (equipotential surface) of the scalar and in the direction of maximum rate of change of constant scalar. Remember this statement when we perform the experiment.

Use the electric potential calculator to determine the electric potential at a point either due to a single point charge or a system of point charges. You can also use this tool to find out the electrical potential difference between two points.

If you want to calculate the electric field due to a point charge, check out the electric field calculator.

Continue reading this article to learn:

What is electric potential?
What is the relation between electric potential and electric potential energy?
How to calculate electric potential?
What is the unit of electric potential?

To understand the idea of electric potential difference, let us consider some charge distribution. This charge distribution will produce an electric field. Now, if we want to move a small charge $q$ between any two points in this field, some work has to be done against the Coulomb force (you can use our Coulomb's law calculator to determine this force). This work done gets stored in the charge in the form of its electric potential energy.

If we consider two arbitrary points, say A and B, then the work done ( $W_{AB}$ ) and the change in the potential energy ( $ΔU\Delta U$ ) when the charge ( $q$ ) moves from A to B can be written as:

$WAB=ΔU=(VA−VB)qW_{AB} = \Delta U = (V_A - V_B)q$ ...... (1)

where $V_A$ and $V_B$ are the electric potentials at A and B, respectively (we will explain what it means in the next section).

If the magnitude of $q$ is unity (we call a positive charge of unit magnitude as a test charge), the equation changes to:

$\Delta V = (V_A - V_B) = \frac{W_{AB}}{q}$ ...... (2)

Using the above equation, we can define the electric potential difference ( $ΔV\Delta V$ ) between the two points (B and A) as the work done to move a test charge from A to B against the electrostatic force.

Remember that the electric potential energy can't be calculated with the standard potential energy formula, $E = m g h$ .

If we take one of the points in the previous section, say point A, at infinity and choose the potential at infinity to be zero, we can modify the electric potential difference formula (equation 2) as:

$V_B = \frac{W_{\infty B}}{q}$

Hence, we can define the electric potential at any point as the amount of work done in moving a test charge from infinity to that point.

We can also define electric potential as the electric potential energy per unit charge, i.e.:

$\frac{\Delta U}{q}$

So you can see that electric potential and electric potential energy are not the same things.

To calculate electric potential at any point A due to a single point charge (see figure 1), we will use the formula:

$V=kqr\scriptsize V = k \frac{q}{r}$

where:

$q$ — Electrostatic charge;
$r$ — Distance between A and the point charge; and
$\frac{1}{4 \pi \epsilon_0}$ — Coulomb's constant.

Fig. 1: Electric potential due to a point charge.

We note that when the charge $q$ is positive, the electric potential is positive. When the charge $q$ is negative electric potential is negative.

Now we will consider a case where there are four point charges, $q_1$ , $q_2$ , $q_3$ , and $q_4$ (see figure 2). The potential at point A due to the charge $q_1$ is:

$V1=kq1r1\scriptsize V_1 = k \frac{q_1}{r_1}$

Fig. 2: Electric potential due to a system of point charges.

We can write similar expressions for the potential at A due to the other charges:

$V2=kq2r2V3=kq3r3V4=kq4r4\scriptsize \begin{align*} V_2 &= k \frac{q_2}{r_2} \\ \\ V_3 &= k \frac{q_3}{r_3} \\ \\ V_4 &= k \frac{q_4}{r_4} \end{align*}$

To get the resultant potential at A, we will use the superposition principle, i.e., we will add the individual potentials:

$V=V1+V2+V3+V4V=k(q1r1+q2r2+q3r3+q4r4)\scriptsize \begin{align*} V &= V_1 + V_2 + V_3 + V_4 \\ \\ V &= k \left (\frac{q_1}{r_1} + \frac{q_2}{r_2} + \frac{q_3}{r_3} + \frac{q_4}{r_4}\right ) \\ \end{align*}$

For a system of $n$ point charges, we can write the resultant potential as:

$V=V1+V2+V3+....+VnV=k(q1r1+q2r2+q3r3+....+qnrn)V=k∑qiri\scriptsize \begin{align*} V &= V_1 + V_2 + V_3 + .... +V_n \\ \\ V &= k \left (\frac{q_1}{r_1} + \frac{q_2}{r_2} + \frac{q_3}{r_3} + .... +\frac{q_n}{r_n}\right ) \\ \\ V & = k \sum \frac{q_i}{r_i} \end{align*}$

In the next section, we will see how to calculate electric potential using a simple example.

Let us calculate the electrostatic potential at a point due to a charge of $\times 10^{-7}\ \rm C$ located at a distance of $cm10\ \rm cm$ .

We are given:

$\times 10^{-7}\ \rm C$ and $10\ \rm cm$ .
Substituting these values in the formula for electric potential due to a point charge, we get:

$\frac{q}{4 \pi \epsilon_0 r}$

$\frac{8.99 \times 10^9\ \rm N \cdot m^2/C^2 \times 4 \times 10^{-7}\ \rm C}{0.1\ m}$

$\times 10^4\ \rm V$
Hence, the electric potential at a point due to a charge of $\times 10^{-7}\ \rm C$ located at a distance of $cm10\ \rm cm$ away is $3.6 \times 10^4\ \rm V$ .

Now we will see how we can solve the same problem using our electric potential calculator:

Using the drop-down menu, choose electric potential due to a point charge.
Enter the value of electric charge, i.e., $4 e - 07$ and the distance between the point charge and the observation point ( $cm10\ \rm cm$ ).
You can also change the value of relative permittivity using Advanced mode.
The calculator will display the value of the electric potential at the observation point, i.e., $3.595 \times 10^4 \ \rm V$ .

The SI unit of electric potential is the volt (V). We can say that the electric potential at a point is 1 V if 1 J of work is done in carrying a positive charge of 1 C from infinity to that point against the electrostatic force.

The unit of potential difference is also the volt. You might be more familiar with voltage instead of the term potential difference. For example, when we talk about a 3 V battery, we simply mean that the potential difference between its two terminals is 3 V.

To write the dimensional formula for electric potential (or electric potential difference), we will first write the equation for electric potential:

$\frac{W}{q}$

Now substituting the dimensional formula for work/energy and charge, we will get the dimensional formula for electric potential as:

$V=[M1L2T−2][AT]=[M1L2T−3A−1]\scriptsize V = \rm \frac{[M^1 L^2T^{-2}]}{[AT]} = [M^1L^2T^{-3}A^{-1}]$

To calculate the electric potential of a point charge (q) at a distance (r), follow the given instructions:

Multiply the charge q by Coulomb's constant.
Divide the value from step 1 by the distance r.
Congrats! You have calculated the electric potential of a point charge.

Yes, electric potential can be negative. The electrostatic potential at a point due to a positive charge is positive. If the charge is negative electric potential is also negative.

The electric potential difference between two points A and B is defined as the work done to move a positive unit charge from A to B. The SI unit of potential difference is volt (V).

Electric potential is a scalar quantity as it has no direction.

Electric potential is the electric potential energy per unit charge. The SI unit of electric potential energy is the joule (J), and that of charge is the coulomb (C). Hence, the SI unit of electric potential is J/C, i.e., the volt (V).

Zero. The electric potential at a point P due to a charge q is inversely proportional to the distance between them. Hence, when the distance is infinite, the electric potential is zero.