With the help of a diagram, obtain an expression for the electric field at a point outside a charged thin spherical shell. The figure shows four positively charged particles. A Gaussian surface encloses q1 and q2 : (i) What is the net electric flux through the Gaussian surface ?

Question

With the help of a diagram, obtain an expression for the electric field at a point outside a charged thin spherical shell. The figure shows four positively charged particles. A Gaussian surface encloses q1 and q2 : (i) What is the net electric flux through the Gaussian surface ? (ii) Which of the particles shown, contribute to the electric field at point P on the surface ?

**OR**

With the help of a diagram, obtain an expression for the electric field of an electric dipole at a point on its equatorial plane. The figure shows an electric dipole placed along X-axis in an external non-uniform electric field, increasing in +x direction. (i) What is the direction of the force experienced by the charge (-q) ? (ii) What is the direction of the net force experienced by the dipole ?

Accepted Answer

![Figure — Deriving the field outside a charged thin spherical shell uses a concentric Gaussian surface at r>R; the NCERT](/figures/electric-charges-and-fields/gaussian-surfaces-point-fig-1-28.svg)

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> [!TLDR]
> Gauss's law gives the field outside a uniformly charged spherical shell as $E=q/(4\pi\varepsilon_0 r^2)$; but note the crucial distinction — the FLUX through a Gaussian surface depends only on the ENCLOSED charge, while the FIELD at any point on that surface is due to ALL charges present, enclosed or not.
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> **Derivation — field outside a charged thin spherical shell:** Consider a thin spherical shell of radius $R$ carrying total charge $q$, uniformly distributed on its surface. To find the field at a point P outside the shell, at distance $r>R$ from the centre, choose a concentric spherical Gaussian surface of radius $r$ through P. By symmetry, $\vec E$ is radial and has the same magnitude everywhere on this Gaussian surface, so
> $$\oint \vec E\cdot d\vec A = E	imes 4\pi r^2$$
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> By Gauss's law, this equals $q_{enc}/\varepsilon_0$. Since the entire shell charge $q$ lies inside the Gaussian surface (as $r>R$),
> $$E	imes 4\pi r^2 = \frac{q}{\varepsilon_0} \implies E=\frac{q}{4\pi\varepsilon_0 r^2}$$
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> This is exactly the field of a point charge $q$ placed at the centre — i.e., outside a uniformly charged spherical shell, the shell behaves as if all its charge were concentrated at the centre.
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> **Applying the same Gauss's-law reasoning to the figure (four point charges q1–q4, Gaussian surface enclosing only q1 and q2):**
> …

Q.With the help of a diagram, obtain an expression for the electric field at a point outside a charged thin spherical shell. The figure shows four positively charged particles. A Gaussian surface encloses q1 and q2 :

Concept understanding — Gauss Law

Why this formula?

Gauss's Law: Why It Holds

1. The Core Idea: Flux as "Flow" of Field

2. The Key Insight: Flux Depends Only on Enclosed Charge

3. Why Shape Doesn't Matter

4. Multiple Charges: Superposition

Q.With the help of a diagram, obtain an expression for the electric field at a point outside a charged thin spherical shell. The figure shows four positively charged particles. A Gaussian surface encloses q1 and q2 :

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