If a spherical surface encloses a uniformly charged sphere, how does the electric field behave at points outside the sphere according to Gauss's Law?

Question

Seekh · Accepted Answer

When the Gaussian surface lies outside the uniformly charged sphere, Gauss’s Law tells us that the total electric flux through the surface depends only on the total charge inside, not on how that charge is distributed. Because the sphere is uniformly charged, the charge inside the surface can be treated as if it were all concentrated at the sphere’s center. The symmetry of a sphere then forces the electric field to be radial and have the same magnitude at every point on the Gaussian surface. Using the relation $E \cdot 4\pi r^{2} = Q/\varepsilon_{0}$, the field at a distance $r$ from the center is $E = \frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r^{2}}$, pointing outward. For example, if the sphere carries a charge of $1\,\mu\text{C}$ and you measure the field at twice the sphere’s radius, the field will be one fourth of its value at the sphere’s surface, just as for a point charge.

If a spherical surface encloses a uniformly charged sphere, how does the electric field behave at points outside the sphere according to Gauss's Law?

Learning Path

Choose the Best Answer

Understanding the Answer

Key Concepts

Deep Dive: Electric Field Effects on Charge

Definition

Topic Definition

Ready to Master More Topics?