Electric Field & Field Lines

The Concept of Electric Field

How does one charge 'know' another charge is nearby without touching it? To explain this 'action at a distance,' Michael Faraday introduced the concept of the Electric Field.

Every charge creates a modification in the space around it. If we bring another charge into this space, it experiences a force. We say the first charge has created an electric field.

Definition:

The electric field $\vec{E}$ at a point is defined as the electrostatic force $\vec{F}$ experienced by a very small unit positive 'test charge' $q_0$ placed at that point, divided by the magnitude of the test charge. $$\vec{E} = \lim_{q_0 \to 0} \frac{\vec{F}}{q_0}$$

• Nature: It is a vector quantity. Its direction is the same as the direction of force on a positive test charge.
• SI Unit: Newton per Coulomb (N/C) or Volt per meter (V/m).
• Dimensions: $[MLT^{-3}A^{-1}]$.

Electric Field due to a Point Charge

Consider a point charge $q$ placed at the origin. To find the field at a distance $r$, we place a test charge $q_0$ there. By Coulomb's Law, the force is: $$F = \frac{1}{4\pi\epsilon_0} \frac{q q_0}{r^2}$$ Dividing by $q_0$, we get the magnitude of the Electric Field: $$E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$$

In vector form: $$\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}$$

Important Note:

1. If $q > 0$ (Positive), the field is radially outwards.
2. If $q < 0$ (Negative), the field is radially inwards.
3. The field $E$ decreases as the square of the distance ($E \propto 1/r^2$).

Electric Field Lines

Electric field lines are a way to visualize the electric field. They are imaginary curves drawn in such a way that the tangent to the curve at any point gives the direction of the electric field at that point.

Properties of Electric Field Lines:

1. Start and End: They start from positive charges and end at negative charges. For a single charge, they start or end at infinity.
2. No Loops: They never form closed loops (unlike magnetic field lines).
3. Continuity: They are continuous curves in a charge-free region.
4. No Intersection: Two field lines never cross each other. Why? Because if they did, the field would have two different directions at the point of intersection, which is physically impossible.
5. Normal to Surface: They are always normal (perpendicular) to the surface of a conductor.
6. Relative Density: The number of lines per unit area (density of lines) represents the strength of the field. Crowded lines mean a strong field; spread-out lines mean a weak field.

Superposition of Electric Fields

Just like forces, electric fields follow the Principle of Superposition. The net electric field at a point due to a system of charges is the vector sum of the fields produced by individual charges at that point. $$\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \dots$$

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🧠 Memory Capsule

• Formula: $E = kQ/r^2$. (Point charge).
• Force on a charge: If a charge $q$ is placed in a field $\vec{E}$, it experiences force $\vec{F} = q\vec{E}$.
• Direction: $\vec{F}$ is in the direction of $\vec{E}$ for positive $q$, and opposite to $\vec{E}$ for negative $q$.
• Uniform Field: A field that has the same magnitude and direction at all points (represented by parallel, equidistant straight lines).
• Tangent Rule: Tangent to a field line = Direction of $\vec{E}$ at that point.

Example 1: Basic Field Calculation

What is the magnitude of the electric field at a point 30 cm away from a point charge of $5 \mu C$ in vacuum?

Solution:

1. Given: $q = 5 \times 10^{-6}$ C, $r = 0.3$ m, $k = 9 \times 10^9$.
2. Formula: $E = k|q|/r^2$.
3. Calculate: $E = (9 \times 10^9) \times (5 \times 10^{-6}) / (0.3)^2$.
4. Simplify: $E = (45 \times 10^3) / 0.09 = 5 \times 10^5$ N/C.
5. Result: The field magnitude is $5 \times 10^5$ N/C.

Example 2: Force in a Field

An electron ($q = -1.6 \times 10^{-19}$ C) is placed in an electric field of $2 \times 10^4$ N/C directed toward the East. Find the magnitude and direction of the force acting on it.

Solution:

1. Formula: $\vec{F} = q\vec{E}$.
2. Magnitude: $F = |q|E = (1.6 \times 10^{-19}) \times (2 \times 10^4) = 3.2 \times 10^{-15}$ N.
3. Direction: Since the electron is negatively charged, the force acts opposite to the field.
4. Result: $3.2 \times 10^{-15}$ N directed toward the West.

Example 3: Field of an Alpha Particle

Find the electric field at a distance of $1 \text{ \AA}$ ($10^{-10}$ m) from an alpha particle.

Solution:

1. Charge of alpha particle: $q = +2e = 3.2 \times 10^{-19}$ C.
2. Distance: $r = 10^{-10}$ m.
3. Apply Formula: $E = (9 \times 10^9 \times 3.2 \times 10^{-19}) / (10^{-10})^2$.
4. Calculate: $E = (28.8 \times 10^{-10}) / 10^{-20} = 2.88 \times 10^{11}$ N/C.

Example 4: Null Point between Two Charges (JEE Level)

Two point charges $q_1 = +4 \mu C$ and $q_2 = +1 \mu C$ are separated by a distance of 30 cm. Find the point on the line joining them where the electric field is zero.

Solution:

1. Concept: At the null point, $E_1 = E_2$. The point must be between the charges because they are like charges.
2. Setup: Let the point be at distance $x$ from $q_1$. Then it is $(30-x)$ from $q_2$.
3. Equation: $k(4)/x^2 = k(1)/(30-x)^2$.
4. Simplify: Take square root: $2/x = 1/(30-x)$.
5. Solve: $60 - 2x = x \Rightarrow 3x = 60 \Rightarrow x = 20$ cm.
6. Result: 20 cm from the $4 \mu C$ charge (or 10 cm from the $1 \mu C$ charge).

Example 5: Net Field at Square Corner

Four equal charges $+q$ are placed at the corners of a square of side $a$. What is the electric field at the center of the square?

Solution:

1. Analyze Symmetry: The distance from each corner to the center is the same ($r = a/\sqrt{2}$).
2. Vector Addition: The field from the top-left charge points toward the bottom-right. The field from the bottom-right charge points toward the top-left. Since charges and distances are equal, they cancel out.
3. Result: By symmetry, the fields from opposite corners cancel each other perfectly. The net electric field at the center is Zero.

Example 6: Suspension in Electric Field

A pith ball of mass 9 mg carries a charge of $5 \mu C$. What must be the magnitude and direction of a vertical electric field required to keep the ball stationary?

Solution:

1. Forces: Gravity ($mg$) downwards, Electric force ($qE$) upwards.
2. Equilibrium: $qE = mg$.
3. Convert Units: $m = 9 \times 10^{-6}$ kg, $g = 9.8$ m/s$^2$, $q = 5 \times 10^{-6}$ C.
4. Calculate: $E = mg/q = (9 \times 10^{-6} \times 9.8) / (5 \times 10^{-6}) = 17.64$ N/C.
5. Direction: Since the charge is positive, the field must be upwards to provide an upward force.

Example 7: Acceleration in a Uniform Field

Find the acceleration of a proton in a uniform electric field of $500$ N/C.

Solution:

1. Identify Constants: $q_p = 1.6 \times 10^{-19}$ C, $m_p = 1.67 \times 10^{-27}$ kg.
2. Force: $F = qE = 1.6 \times 10^{-19} \times 500 = 8 \times 10^{-17}$ N.
3. Acceleration: $a = F/m = (8 \times 10^{-17}) / (1.67 \times 10^{-27})$.
4. Result: $a \approx 4.79 \times 10^{10}$ m/s$^2$.

Example 8: Field at a Distance r >> a (Dipole Preview)

Two charges $+q$ and $-q$ are separated by a small distance $2a$. Find the electric field at a point on the perpendicular bisector at a very large distance $r$.

Solution:

1. Geometry: The fields from $+q$ and $-q$ have equal magnitude $E = kq/(r^2 + a^2)$.
2. Components: Vertical components cancel; horizontal components add up.
3. Calculation: $E_{net} = 2E \cos \theta = 2 [kq/(r^2+a^2)] \times [a/\sqrt{r^2+a^2}] = k(2qa)/(r^2+a^2)^{3/2}$.
4. Approximation: If $r \gg a$, then $(r^2+a^2)^{3/2} \approx r^3$.
5. Result: $E \approx k(2qa)/r^3$.

Example 9: Superposition on Triangle (Midpoint)

Two charges $+q$ and $+q$ are at the base corners of an equilateral triangle of side $L$. Find the field at the midpoint of the base.

Solution:

1. Analyze base midpoint: The two charges are equidistant ($L/2$) from the midpoint but on opposite sides.
2. Vector Sum: The field from the left charge points right. The field from the right charge points left.
3. Result: Since magnitudes are equal ($kQ/(L/2)^2$), they cancel out. $E_{net} = 0$.

Example 10: Deflection of a Particle

A charged particle enters a uniform electric field $\vec{E}$ with velocity $\vec{v}$ perpendicular to the field. Describe its path.

Solution:

1. Forces: The force $qE$ acts only in one dimension (say y-axis). There is no force in the x-axis ($v_x$ is constant).
2. Motion: This is analogous to projectile motion where gravity acts in one direction.
3. Result: The path of the charged particle in a uniform electric field is a Parabola.