A positive charge Q = 87 nC is uniformly distributed over the volume of a sphere of radius R = 4.60 cm. Suppose a spherical cavity of radius R/2 is cut out of the solid sphere, the... A positive charge Q = 87 nC is uniformly distributed over the volume of a sphere of radius R = 4.60 cm. Suppose a spherical cavity of radius R/2 is cut out of the solid sphere, the center of the cavity being a distance R/2 from the center of the original solid sphere. The cut-out material and its charge are discarded. What is the magnitude of the electric field produced by this new charge distribution at point P, a distance r = 0.32 m from the center of the original sphere?
Understand the Problem
The question is asking us to calculate the magnitude of the electric field produced by a new charge distribution after removing a cavity from a uniformly charged sphere. We will use principles of electrostatics and the superposition of electric fields to find the solution.
Answer
The electric field inside the cavity is $E_{\text{cavity}} = 0$. The electric field outside is given by $E = \frac{Q}{4 \pi \epsilon_0 d^2}$.
Answer for screen readers
The electric field inside the cavity is $E_{\text{cavity}} = 0$. Outside the sphere, it can be calculated as $E = \frac{Q}{4 \pi \epsilon_0 d^2}$.
Steps to Solve
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Understand the Charge Distribution The original uniformly charged sphere has charge density $\rho$ and radius $R$. When we remove a spherical cavity of radius $r$ (and centered inside the sphere), we can find the effective electric field caused by the remaining charge distribution.
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Calculate the Total Electric Field To find the electric field $E$ outside the sphere, we need to treat it as if the entire sphere (with radius $R$) is charged. The total charge $Q$ of the sphere can be computed as: $$ Q = \rho \cdot \frac{4}{3} \pi R^3 $$
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Determine the Electric Field Outside the Sphere For a point outside the sphere at distance $d$ from the center (where $d > R$), the electric field can be calculated using Gauss's law: $$ E = \frac{Q}{4 \pi \epsilon_0 d^2} $$ This means the field behaves like a point charge at the origin.
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Electric Field Inside the Cavity Inside the cavity, the electric field will depend on the symmetry and the absence of charge. By Gauss's law, the field in the cavity will be zero if it's a conducting material or if the cavity is centered symmetrically. If the cavity is not affecting the uniformity, we have: $$ E_{\text{cavity}} = 0 $$
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Final Electric Field Calculation Hence, if we want the electric field at a point inside the remaining charged sphere (not including the cavity), you can still consider the field due to the remaining charge distribution, which constructs a complete uniform field field considering the cavity is effectively "filled by nothing".
The electric field inside the cavity is $E_{\text{cavity}} = 0$. Outside the sphere, it can be calculated as $E = \frac{Q}{4 \pi \epsilon_0 d^2}$.
More Information
When calculating electric fields, it’s important to note that the symmetry of the charge distribution plays a critical role, especially when dealing with cavities. The concept of superposition helps simplify these conditions, allowing us to ignore the internal cavity in certain cases.
Tips
- Forgetting to consider the symmetry of the cavity can lead to incorrect assumptions about the electric field.
- Not applying Gauss’s law correctly when determining fields around charged distributions can result in incorrect electric field calculations.
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