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  • Figure shows a solid hemisphere with a charge of 5 nC distributed uniformly throughout its volume. The hemisphere lies on a plane and point P is located on the plane, along a radial line from the c

Figure shows a solid hemisphere with a charge of 5 nC distributed uniformly throughout its volume. The hemisphere lies on a plane and point P is located on the plane, along a radial line from the centre of curvature at distance 15 cm. The electric potential (in Volt)  at point P due to the hemisphere, is :

Option: 1 150

Option: 2 300

Option: 3 450

Option: 4 600

Answers (1)

best_answer

 

Outside the sphere (P lies outside the sphere) -

\dpi{100} E_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r^{2}}=\frac{\sigma R^{2}}{\epsilon _{0}r^{2}}

V_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r}=\frac{\sigma R^{2}}{\epsilon _{0}r}

 

- wherein

\sigma - surface charge density.

 

 

 By argument of symmetry, it will be half of the potential produced by the full sphere

 Q Charge on hemisphere = Q,

 so charge on sphere = 2Q

\Rightarrow \frac{1}{2}. \frac{ K(2Q)}{R} = \frac{KQ}{R}

V = \frac{KQ}{R} = \frac{9\times 10^{9}\times 5\times 10^{-9}}{15\times 10^{-2}} = 300 V

Posted by

Ritika Kankaria

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