Force Due to Continuous Charge Distribution

1. Statement of the Concept / Overview

When charge is distributed over a line, surface, or volume, the total electric force on a test charge is obtained by breaking the distribution into infinitesimally small charge elements and integrating the forces due to these elements.

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2. Clear Explanation and Mathematical Derivation

For a continuous distribution, replace discrete charges (q_i) with infinitesimal charge elements [dq].

Force on a test charge [q_0] due to [dq] at distance [r]:

[d\vec{F}] [= k \dfrac{q_0,dq}{r^2} \hat{r}]

Total force:

[\vec{F}] [= \displaystyle \int d\vec{F}]

The form of [dq] depends on the type of distribution:

(1) Linear Charge Distribution

If [\lambda] = charge per unit length:

[
dq = \lambda dl
]

(2) Surface Charge Distribution

If [\sigma] = charge per unit area:

[
dq = \sigma dA
]

(3) Volume Charge Distribution

If [\rho] = charge per unit volume:

[
dq = \rho dV
]

Then integrate appropriately based on geometry and limits.

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3. Dimensions and Units

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4. Key Features

• Uses integration instead of simple Coulomb summation.
• Direction of force requires symmetry considerations.
• Eliminates the need to treat large bodies as point charges.
• Applicable to rods, rings, discs, cylinders, spheres, sheets, etc.
• Superposition principle is inherently built into the method.
• Continuous distribution is essential for real-world physics (e.g., capacitor plates, charged rods).

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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: Force on axis of a charged ring

A ring of radius [R] carries total charge [Q]. Find the force on a charge [q_0] placed on its axis at distance [x].

Solution:

1. Take small element [dq] on the ring.
2. Force due to [dq]:
   [dF] [= k \dfrac{q_0,dq}{(R^2 + x^2)}]
3. Radial components cancel; only axial component remains:
   [
   dF_x = dF \cos\theta
   ]
4. From geometry:
   [\cos\theta] [= \dfrac{x}{\sqrt{R^2 + x^2}}]
5. Integrate:
   [F_x] [= \displaystyle \int dF_x] [= k q_0 x\dfrac{1}{(R^2+x^2)^{3/2}} \int dq]
6. (\int dq = Q)

Final Answer:
[F] [= k \dfrac{q_0 Q x}{(R^2 + x^2)^{3/2}}]

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Question 2: Force due to a uniformly charged rod

A rod of length [L] has linear charge density [\lambda]. A charge [q_0] is placed at distance [a] from one end along its axis. Find the force.

Solution:

1. Take element at distance [x];
   [
   dq = \lambda dx
   ]
2. Distance from charge = [(a + x)]
   Force:
   [dF] [= k\dfrac{q_0 \lambda dx}{(a+x)^2}]
3. Integrate from [0] to [L]:

[F] $= k q_0 \lambda \left[ -\dfrac{1}{a+x} \right]_0^L$

4. Solve:

[F] $= k q_0 \lambda \left( \dfrac{1}{a} – \dfrac{1}{a+L} \right)$

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Question 3: Force due to infinite line charge

Infinite line charge with density [\lambda]; find the force on charge [q_0] at perpendicular distance [r].

Use standard result for electric field:

[
E = \dfrac{\lambda}{2\pi\epsilon_0 r}
]

Thus:

[F = q_0 E] [= \dfrac{q_0 \lambda}{2\pi\epsilon_0 r}]

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Question 4: Force due to a uniformly charged disc

Axial field of disc is:

[E] $= \dfrac{\sigma}{2\epsilon_0} \left( 1 – \dfrac{x}{\sqrt{x^2 + R^2}} \right)$

Hence:

[
F = q_0 E
]

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Question 5: Force due to a spherical shell (outside)

A spherical shell carries charge [Q]. A charge [q_0] is at distance [r > R].

Shell behaves like a point charge:

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