When we study a charge or a system of charge, we see that every charge has a definite amount of electric field which exerts the electrostatic force of repulsion or attraction on other charges and charged particles according to their nature. The effect of electricity can be experienced with the help of a unit charge, also known as a test charge. The electric field around a charge or system of charge can be described in two forms; The electric field and The electric potential.
The electric potential of a dipole is the electrostatic potential that arises due to a dipole. Now, first of all, we should clarify what is electric potential? In simple words, we can say that the electric potential for a point in an electric field is the piece of work we have to do for moving a point positive charge of magnitude one coulomb from infinity position to that considered point in which the effect of the electric field is present which continues exerting electrostatic force. Hence, electric potential can be expressed as work done on a unit charge.
The standard measurement unit of Electric potential is Volt. Hence, one-volt electric potential point is the one joule of work that has been done to bring a unit positive charge from an infinite far position to a target point under the influence of force due to an electric field.
$$\mathrm{V=\frac{W}{q}}$$
The electric dipole is a system of two charges having the same value and opposite nature and these are connected by a line that passes through their center. The length of the line that passes through its center is called dipole length.
To measure the strength of electric dipole we use the term dipole moment. A Dipole moment can be defined as a vector whose magnitude is the product of charges and the total separation between them and the direction of the vector will be along the axis from negative to positive. It is denoted by p.
The SI unit of dipole moment is coulomb metre.
On the other hand, if we use a system of charges having more than 2 charges then the electric potential is called the electric potential developed by the system of charges.
The electric potential created at any point, such as P because of an electric dipole can be expressed for two points; axial point and equatorial point.
Electric Potential developed by a dipole at an axial point
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In the above diagram, there is a dipole having pair of charges; positive at A and negative at B, separated by some distance denoted by d which is equal to 2a. There is point P which lies on the axis of the dipole which is r distance away from the center of the dipole.
Now, the electric potential for point P by both charges is
$$\mathrm{V=V_{1}+V_{2}}$$
$$\mathrm{V_{1}=\frac{1}{4\pi \epsilon _{0}}.\frac{-q}{AP}}$$
Now,
$$\mathrm{V_{2}=\frac{1}{4\pi \epsilon _{0}}.\frac{+q}{BP}}$$
Total Potential at P,$\mathrm{V=V_{1}+V_{2}}$
So,
$$\mathrm{V=\frac{1}{4\pi \epsilon _{0}}.\frac{-q}{AP}+\frac{1}{4\pi \epsilon _{0}}.\frac{+q}{BP}}$$
We know that $\mathrm{AP=r+\frac{d}{2}}$ and $\mathrm{BP=r-\frac{d}{2}}$ whereas $\mathrm{\frac{d}{2}=a}$.
$$\mathrm{Thus,V=\frac{1}{4\pi \epsilon _{0}}.\frac{-q}{r+a}+\frac{1}{4\pi \epsilon _{0}}.\frac{+q}{r-a}}$$
$$\mathrm{V=\frac{q}{4\pi \epsilon _{0}}.\left\{\frac{1}{r-a}-\frac{1}{r+a} \right\}}$$
$$\mathrm{V=\frac{q}{4\pi \epsilon _{0}}.\left\{\frac{(r+a)-(r-a)}{r^{2}-a^{2}} \right\}}$$
$$\mathrm{V=\frac{q}{4\pi \epsilon _{0}}.\left\{\frac{q\times 2a}{r^{2}-a^{2}} \right\}}$$
However, $\mathrm{p=q\times 2a}$
Hence,$\mathrm{V=\frac{q}{4\pi \epsilon _{0}}.\left\{\frac{p}{r^{2}-a^{2}} \right\}}$
If the dipole is very small then $\mathrm{a^{2}<<r^{2}}$
So,
$$\mathrm{V=\frac{q}{4\pi \epsilon _{0}}.\frac{p}{r^{2}}}$$
Electric Potential at an equatorial point of the dipole
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Let a dipole of a pair of charges having opposite nature and equal magnitude, which is -q and +q, both charges are 2l distant from each other. There is a random point P which lies position linear to the perpendicular bisector of the dipole and at a distance of P is r from the center of the dipole.
Now, the distance of point P from -q and +q is the same which is $\mathrm{\sqrt{r^{2}+l^{2}}}$
According to the formula of the electric potential created at position P because of both charges is
$$\mathrm{V=V_{1}+V_{2}}$$
$$\mathrm{V=\frac{1}{4\pi \epsilon _{0}}.\frac{-q}{AP}+\frac{1}{4\pi \epsilon _{0}}.\frac{+q}{BP}}$$
Here, both AP and BP are the same.
$$\mathrm{So,V=\frac{1}{4\pi \epsilon _{0}}.\frac{-q}{\sqrt{r^{2}+l^{2}}}+\frac{1}{4\pi \epsilon _{0}}.\frac{+q}{\sqrt{r^{2}+l^{2}}}}$$
$$\mathrm{Thus,-\frac{1}{4\pi \epsilon _{0}}.\frac{q}{\sqrt{r^{2}+l^{2}}}+\frac{1}{4\pi \epsilon _{0}}.\frac{q}{\sqrt{r^{2}+l^{2}}}=0}$$
$$\mathrm{V=0}$$
Therefore, from the above expression find out that the electric potential at the equatorial point of an electric dipole is zero.
The electric potential created by a dipole at the point which is linear to the dipole is $\mathrm{V=\frac{q}{4\pi \epsilon _{0}}.\left\{\frac{p}{r^{2}-a^{2}} \right\}}$ whereas the electric potential at the equatorial point of an electric dipole is zero.
Q1. Why electric potential at an equatorial point of a dipole is zero?
Ans. Yes, the equatorial point of a dipole indeed has zero electric potential because the point is in between both charges which have equal value and opposite nature. Therefore, both charges cancel the effect on each other and the result becomes zero.
Q2. What is an Electric Field?
Ans. An electric field is a region around any charged body or particle in which we can feel the force of attraction or repulsion when we bring a test charge to that region. This field is directly proportional to the magnitude of charge, if a body has a large value of charge then the field will be large and strong and vice versa.
Q3. What is the strength of the electric field?
Ans. we can define the electric field strength as the force exerted by a charge in an electric field on a particular point. If there is a system of charge then the strength will be the sum of force exerted by all charges.
Q4. How is the Electric field has a relation to Electric Potential?
Ans. The Electric field and Electric Potential have a direct relationship which is given below −
$$\mathrm{E=-\frac{dV}{dr}}$$
Q5. What will be the value of the electric field at a point where electric potential remains constant?
Ans. According to the relation, $\mathrm{E=-\frac{dV}{dr}}$, we get that the derivative of electric potential to distance is also related to the electric field.
Hence, if the electric potential becomes constant at any point then at that point the electric field will be zero.
Q6. What are the direction, SI unit, and dimension formula of the Electric Dipole Moment?
Ans. The dipole moment is a vector quantity and lies along the axis of the dipole from negative to positive. Moreover, Coulomb Meter is the SI unit and Dimension formula of $\mathrm{M^{0}L^{1}T^{1}I^{1}}$
Q7. What do you mean by quantization of charge?
Ans. The quantum represents the minimum amount of any object. Here, quantization of charge means that the charge on a body will always be an integral multiple of its basic quantum.
$$\mathrm{q=ne}$$
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