Electric Field

1. Statement of the Concept / Overview

The electric field at a point in space is defined as the force experienced per unit positive test charge placed at that point, without disturbing the existing charge configuration.

[
\vec{E} = \dfrac{\vec{F}}{q_0}
]

It is a vector quantity and represents the influence a charge exerts on the surrounding space.

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2. Clear Explanation and Mathematical Derivation

Electric Field Due to a Single Point Charge

Consider a point charge [Q]. Place a test charge [q_0] at distance [r], which experiences force:

[\vec{F} = k \dfrac{Q q_0}{r^2} \hat{r}]

Electric field:

[
\vec{E} = \dfrac{\vec{F}}{q_0}
]

Substituting:

[
\vec{E} = k \dfrac{Q}{r^2} \hat{r}
]

Thus:

• If [Q > 0], field is outward (radial).
• If [Q < 0], field is inward.

 

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Principle of Superposition

For charges [Q_1, Q_2, \dots]:

[\vec{E}_{\text{net}}] [= \vec{E}_1 + \vec{E}_2 + \dots]

Because electric field is vectorial.

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Electric Field Lines

• Start on positive charges, end on negative charges.
• Never intersect.
• Tangent at any point gives direction of [\vec{E}].

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3. Dimensions and Units

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4. Key Features

• Vector quantity
• Defined independent of test charge value
• Represents intensity of electric influence
• Obeys superposition principle
• Determines motion of charges
• Originates from positive charges, terminates on negative charges
• Infinite range (due to [1/r^2] law)

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5. Important Formulas to Remember

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6. Conceptual Questions with Solutions

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7. FAQ / Common Misconceptions

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8. Practice Questions (With Step-by-Step Solutions)

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Question 1

Find the electric field due to a point charge [Q = 5,\text{μC}] at a distance of [0.2,\text{m}].

Step-by-step solution:

1. Use
   [
   E = k\dfrac{Q}{r^2}
   ]
2. Substitute:
   [
   E = 9\times10^9 \cdot \dfrac{5\times10^{-6}}{(0.2)^2}
   ]
3. Compute denominator: [(0.2)^2 = 0.04]

[
E = 9\times10^9 \cdot \dfrac{5\times10^{-6}}{0.04}
]
5.
[
E = 9\times10^9 \cdot 1.25\times10^{-4}
]
6.
[
E = 1.125\times10^6\ \text{N/C}
]

Final Answer:
[
1.125\times10^6\ \text{N/C}
]

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Question 2

Two charges [+4,\text{μC}] and [-2,\text{μC}] are placed 1 m apart. Find the field at the midpoint.

Solution:

1. Distance from midpoint = [0.5] m
2. Field due to [+4] μC is outward; due to [-2] μC is toward the charge.
3. Both point in the same direction at midpoint.

[
E_1 = k\dfrac{4\times10^{-6}}{(0.5)^2}
]
[
E_2 = k\dfrac{2\times10^{-6}}{(0.5)^2}
]

[
E_{\text{net}} = (E_1 + E_2)
]

[
E_{\text{net}}] [= 9\times10^9 \left( \dfrac{6\times10^{-6}}{0.25} \right)
]

[
E_{\text{net}}] [= 9\times10^9 \times 24\times10^{-6}
]

[
E_{\text{net}} = 216000\ \text{N/C}
]

Final Answer:
[
2.16\times10^5\ \text{N/C}
]

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Question 3

Find the direction of field at the midpoint between two equal positive charges.

Solution:

• Both charges push the test charge away.
• Directions are opposite at midpoint.
• Magnitudes equal → cancel.

Final Answer:
[
\vec{E}_{\text{mid}} = 0
]

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Question 4

A charge of [-3,\text{μC}] is placed in a region where the electric field is [500,\text{N/C}] rightward. Find the force.

[
F = qE = (-3\times10^{-6})(500)
]
[
F = -1.5\times10^{-3}\ \text{N}
]

Direction: Leftward (opposite to field).

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Question 5

Electric field at a point is [200,\text{N/C}]. What force acts on a charge [q = 0.01,\text{C}]?

[F = qE] [= 0.01 \times 200] [= 2\ \text{N}]

Answer:
[
2\ \text{N}
]