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A solid metallic sphere has charge $+3Q$. Concentric with this sphere is a conducting spherical shell having charge $-Q$. The radius of the sphere is $a$ and that of the spherical shell is $b$ $(b>a)$. What is the electric field at a distance $R$ such that $a<R<b$ from the centre?

  1. $\dfrac{Q}{2\pi\varepsilon_0 R^2}$
  2. $\dfrac{3Q}{2\pi\varepsilon_0 R^2}$
  3. $\dfrac{3Q}{4\pi\varepsilon_0 R^2}$
  4. $\dfrac{4Q}{4\pi\varepsilon_0 R^2}$

Correct answer: $\dfrac{3Q}{4\pi\varepsilon_0 R^2}$

Solution

For $a<R<b$, a Gaussian surface encloses only the charge on the inner metallic sphere, which is $+3Q$. The charge on the outer shell does not affect the field inside its material region. Using Gauss’s law gives $E\cdot 4\pi R^2=3Q/\varepsilon_0$.

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