Relation between Electric Intensity and Potential Gradient, Action of Electric field on a Charged Particle and Equipotential

Relation between Electric Intensity and Potential Gradient

The change of electric potential with respect to distance is called potential gradient. It is denoted by dv/dx.

The value of electric potential (V) as electric field intensity (E) at distance ‘x’ from the charge ‘Q’ is given by

$$V = \frac {1}{4\pi \epsilon _o}\frac Qx \dots (i)$$

$$E = \frac {1}{4\pi \epsilon _o}\frac {Q}{x^2} \dots (i)$$

differentiating equation (i) with respect to 'x' we get,

$$\frac {d(v)}{dx} = \frac {d}{dx} \left ( \frac {1}{4 \pi \epsilon _o} \frac Qx \right ) $$

$$= \frac {Q}{4\pi \epsilon_o}\frac {d}{dx} (x^{-1}) $$

$$= \frac {Q}{4\pi \epsilon_o} {(-1)}{x^{-1-1}} $$

$$ = -\left (\frac {Q}{4\pi \epsilon_o} \frac {1}{x^2} \right )$$

$$= -E$$

$$\boxed {\therefore E = - \frac {dV}{dx}}$$

hence, the negative of potential gradient is equal with electric field intensity.

Acton of Electric field on a Charged Particle

When a charged particle is placed in an electric field, it will experience a force which is given by

$$F = QE$$

where E is the electric field strength and Q is the charge on the particle.

Let us suppose that an electric field is applied between two parallel plates X and Y kept at a distance d apart as shown in the figure. If the potential difference across the plate iv V, then electric field strength, \(E = \frac Vd \).

Suppose, an oil drop of charge Q, radius rand density \(\rho\) is introduced in the electric field between two parallel plates, then force on oil drop due to electric field.

$$F_e = QE$$

Weight of the oil drop, \(W = mg\)

$$= V\rho g $$

$$= \frac 43 \pi r^3 \rho g$$

where \(V = \frac 43 \pi r^3 \) is the volume of the spherical oil drop and g is the acceleration due to gravity.

As the electron on the oil drop has negative charge, the oil drop will experience force in upward direction.

For the oil drop to remain stationary between the plates, the force on oil drop due to the electric field must be equal to the weight of the oil drop,

$$F_e = W$$

$$\text {or,} QE = mg $$

$$\text {or,} Q.\frac Vd = \frac 43 \pi r^3 \rho g $$

$$Q = \frac {4 \pi r^3 \rho gd}{3V} $$

which gives the magnitude of charge on the oil drop to remain stationary between two parallel plates. If a number of electrons are present in the oil drop, then charge Q on it is ne, where e is the charge of the electron. The number of electrons attached to the oil drop is obtained as

$$ne = \frac {4 \pi r^3 \rho gd}{3V} $$

$$n = \frac {4 \pi r^3 \rho gd}{3eV} $$

where, \(e = 1.6 \times 10 ^{-19}\) is the charge of an electron.

Equipotential Surface

Any surface over which the potential is same is called an equipotential surface. The potential difference between any two points on an equipotential surface is zero.

The potential difference between two points A andB is

$$V_B - V_A = W_{AB} $$

If the points A and B lies on the equipotential surface, then

$$V_B - V_A = 0 $$

$$\therefore V_A = V_B$$

$$\therefore W_{AB} = 0 $$

This means no work is done in moving the test charge from one point to the other point on an equipotential surface. It is possible only if the test charge does not experience any force along the equipotential surface. However, the motion of charge is always perpendicular to the electrostatic surface, when the charge moves on the equipotential surface. Hence, the lines of force, which start normally from a charged conductor cut an equipotential surface everywhere at right angles.

Properties of Equipotential Surfaces

The equipotential surfaces have following properties:

1. Equipotential surfaces are always perpendicular to the electric lines of force.
2. No work is done in moving a test charge from one point to the other points on an equipotential surface.
3. Two equipotential surfaces never intersect each other. If they do so, at the point of intersection there will be two values of the potentials which is impossible.
4. Equipotential surface indicates regions of strong and weak electric fields.

Potential Due to Several charges

The net electric potential at a point due to a number of charges is equal with the algebraic sum of electric potential charges.

If V₁, V₂, V₃...... V_n be the individual electric potential produced at point 'P'. Separately, by the charges Q₁, Q₂, Q₃.....Q_n at distance r₁, r₂, r₃, .....r_n respectively. Then the total electric potential 'U' is given by

$$V = V_1 + V_2 + V_3 \dots V_n$$

$$= \frac {1}{4\pi \epsilon _o} \frac {Q_1}{r_1} + \frac {1}{4\pi \epsilon _o} \frac {Q_2}{r_2} + \frac {1}{4\pi \epsilon _o} \frac {Q_3}{r_3} + \dots + \frac {1}{4\pi \epsilon _o} \frac {Q_n}{r_n}$$

$$\boxed {V = \frac {1}{4\pi \epsilon _o} \left ( \frac {Q_1}{r_1} + \frac {Q_2}{r_2} + \frac {Q_3}{r_3} + \dots \frac {Q_n}{r_n} \right )} $$