TS EAMCET · Physics · Electrostatics
The electrostatic potential inside a charged sphere is given as \(V=A r^2+B\), where \(r\) is the distance from the centre of the sphere, \(A\) and \(B\) are constants. Then, the charge density in the sphere is
- A \(16 A \varepsilon_0\)
- B \(-6 A \varepsilon_0\)
- C \(20 A \varepsilon_0\)
- D \(-15 A \varepsilon_0\)
Answer & Solution
Correct Answer
(B) \(-6 A \varepsilon_0\)
Step-by-step Solution
Detailed explanation
Let the volume charge density be \(\rho\). The electric field inside the charged sphere, \( E=\frac{K q r}{R^3} \) where, \(R=\) Radius of charged sphere \(r=\) Distance of the point \((P)\) inside the sphere As volume charge density is \(\rho\), then in terms of charge density,…
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