TS EAMCET · Physics · Electrostatics
The electrostatic potential inside a charged spherical ball is given by \(\Phi=a r^2+b\), where \(r\) is the distance from the centre and \(a, b\) are constant. Then, the charge density inside the ball is ( \(\varepsilon_0=\) permittivity in free space).
- A \(-6 a \varepsilon_0 r\)
- B \(-6 a \varepsilon_0\)
- C \(-24 \pi a \varepsilon_0\)
- D \(-24 \pi a \varepsilon_0 r\)
Answer & Solution
Correct Answer
(B) \(-6 a \varepsilon_0\)
Step-by-step Solution
Detailed explanation
Potential function given is \(\phi=a r^2+b\) Electric field in a region is…
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