Chapter-3 ( Electric Potential )

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Chapter-3(Electric Potential)

Electric Potential: The electrical state for which flow of charge between two charged bodies takes place is called electric potential. In other word, the electric or electrostatic potential at any point of an electric field is defined as potential energy per unit charge at that point. Electric potential is represented by letter V. That is,V=U/q' or U=q'V…………………(1)Electric potential is a scalar quantity since both charge and potential energy are scalar quantities. S.I. unit of electric potential is Volt which is equal to Joule per Coulomb. In other word, The amount of work done in bringing a unit positive charge from infinity to a point in electric field is called electric potential at that point.

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If W is the work done in bringing a positive charge q from infinity to a point in an electric field, then potential at that point, according to definition is V = W/q. Potential V has a sign. Potential at a point can be positive or negative. If the electric field is due to positive charge, i.e. if the field is positive, then in order to bring a unit positive charge from infinity work is to be done against the repulsive force of the similar charges. So, potential due to a positive charge is positive. But if the field is due to negative charge, no work is to be done by external force to bring a unit positive charge from infinity. The attractive force will act between this positive and negative charge. So, potential due to a negative charge is negative.

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Potential difference: The difference of potentials at two points in an electric field is called potential difference. Or, The amount of work done in transferring a unit positive charge from one point to another point in an electric field is called potential difference between the two points.Let the potentials at two points A and B in an electric field is VA and VB respectively. Then

VB - VA = ∆V = WAB/q0 = W/q0.

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Potential and Electric Field Strength:

Fig-1 shows two points A and B in a uniform electric field E. Let the distance of A from B is d. We assume that a positive test charge qo is being moved by an external agent from A to B. The electric force on the charge is qoE and points downward. To move the charge upward , the force on qo must be countered by an external force F of the same magnitude but directed upward.

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The work done by the agent that supplies this force is WAB = F × d = qoEd……………..(1)If the electric potential difference between the points A and B is VB - VA, then we can write,VB - VA = WAB/q0 = qoEd/ q0= Ed…………….(2)Eqn. (2) gives the relation connecting potential difference and field strength for a simple special case. From eqn. (2) it appears that another unit for E is volt per meter. It can be proved that volt per meter is identical with Newton per coulomb, or, 1 V/m = 1 N/C.

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In Fig-1 B is at a higher potential than A because external agent would have to do positive work to push a positive charge from A to B against the direction of the field.Let us now investigate the relation between V and E in the case in which the field is not uniform and in which the test body moves along a path which is not straight. The electric field exerts a force qoE on the test charge as shown in Fig-2. To keep the test charge from accelerating i.e., if the test charge is to move with a constant velocity, the external agent must apply a force F = -qoE for all positions of the test charge.If the external agent causes the test charge to move through a displacement dl along a path from A to B, the element of work done by the external agent is Fcosθ dl = F . dl, where Fcosθ is the component of the force in the direction of displacement.

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To find the total work WAB done by the external agent in moving the test charge from A to B, we integrate (add up) the work contributions for all infinitesimal segments into which the path is divided. This lead to

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If the point A is at infinite distance then the potential VA at infinity is taken as zero. Then, eqn. (4) gives the potential V at the point B. By dropping the subscript B,

Eqns. (4) and (5) allow us to calculate either the potential difference between ant two points or the potential at any point if E is known at various points in the field.

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Example-1: Let a test charge be moved without acceleration from A to B over the path as shown in Fig-2. Compute the potential difference between A and B.

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Electric potential due to a point charge: Fig-3 shows two points A and B near an isolated point charge q. For simplicity we assume that A, B and q lie on a straight line. To compute the potential difference between points A and B, we assume that attest charge qo is moved without acceleration along a radial line from A to B. The potential difference between the points A and B is given by,

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As we move a distance dl to the left, we are moving in the direction of decreasing r because r is measured from q as an origin. Thus, dl = - dr. Therefore, E . dl = - E dl = E dr. Putting the value of E . dl in eqn. (1), we get,

But we know,

where r is the distance of the point from the charge q. Using this in eqn. (2), we have,

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Let the reference point A be at infinity (i.e., rA→ ∞). Then VA = 0 at this position, and dropping the subscript B, we obtain,

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Electric potential due to collection of charges: For a collection of charges, the potential at a point is calculated due to each individual charge, as if the other charges were not present. These potentials are then added and we obtain,

Where r1 is the distance of the point from the charge q1, r2 from q2 and r3 from q3, etc. Thus,

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If the charge distribution is continuous, rather than being a collection of points, the sum in eqn. (2) must be replaced by an integral. That is,

Where dq is a differential element of the charge distribution, r is its distance from the point at which V is to be calculated and dV is the potential it establishes at that point.

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Example-2: What is the potential at the center of the square if Fig-4?

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Example-3: What must the magnitude of an isolated positive point charge be for the electric potential at 10 cm from the charge to be +100 Volts?

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Example-4: Three charges are placed at three corners of a square as shown in Fig-5. Find the potential at point P.

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Two equal charges q of opposite sign, separated by a distance 2a, constitute an electric dipole. The electric dipole moment p has the magnitude 2aq and points from negative charge to the positive charge. Fig-6 shows such a dipole.

Electric potential due to electric dipole

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Here we would like to derive an expression for the electric potential V at any point of space due to a dipole, provided only that the point is not too close to the dipole. Let the point P where we would like to calculate the potential. We know,

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Using eqn. (2) in eqn. (1), we get

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Eqn. (3) gives the exact value of the potential at the point P. But for an ideal dipole (2a << r) r1 →r, r2 →r and ,and in the limit,

The quantity 2aq is called the electric dipole moment, p. Hence eqn. (4) reduces to,

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Special cases: (1) when the point P lies on the axial line of the dipole on the side of the positive charge,

(2) when the point P lies on the axial line of the dipole on the side of the negative charge,

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(3) when the point P lies on the equatorial line of the dipole, θ = 900 so, cosθ = 0, so from eqn. (5), VP = 0…………..(8). This equation reflects the fact that no work is done in bringing a charge from infinity to the dipole along the perpendicular bisector of the dipole.

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Electric potential energy: Fig-7 shows two charges q1 and q2 at a distance r apart. If the separation between them is increased, work must be done by an external agent. The work will be positive if the charges are opposite in sign and negative otherwise. The energy represented by this work can be thought of as stored in the system q1 + q2 as electric potential energy. Like all forms of potential energy, this energy can also be transformed into other forms.

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The electric potential energy of a system of point charges may be defined as the work required to assemble this system of charges by bringing them in from an infinite distance. We assume that the charges are all at rest when they are infinitely separated, that is they have no initial kinetic energy.In Fig-7 let us imagine q2 removed to infinity and at rest. The electric potential at the original site of q2 due to q1 is given by,

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If q2 is now moved from infinity to its original distance r from q1, the work required is, W = Vq2

[As, V = W/q]

This work is precisely the electric potential energy U of the system q1 + q2. Thus,

The subscript of r emphasizes that the distance involved is in between point charges q1 and q2.

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Example-5: Two protons in a nucleus of U238 are 6.0× 10-15 m apart. What is their mutual electric potential energy?

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Example-6: Three charges are arranged as in the figure. What is their mutual potential energy? Assume that q = 1.0 × 10-7 C and a = 10cm.

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Chapter-3 ( Electric Potential )