Force on one Hemisphere by another!

Consider a uniformly charged solid dielectric sphere of total charge Q Q and radius R R . The force exerted by one hemisphere on other hemisphere is?

3 Q 2 64 π ϵ 0 R 2 \frac{3Q^2}{64 \pi \epsilon_0 R^2} 3 Q 2 16 π ϵ 0 R 2 \frac{3Q^2}{16 \pi \epsilon_0 R^2} Q 2 36 π ϵ 0 R 2 \frac{Q^2}{36 \pi \epsilon_0 R^2} Q 2 9 π ϵ 0 R 2 \frac{Q^2}{9 \pi \epsilon_0 R^2}

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1 solution

Nishant Rai
May 14, 2015

Not able to understand the last step of integration

Kyle Finch - 6 years, 1 month ago

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d ( R 2 x 2 ) = 2 x d x d(R^2 - x^2) = -2x~dx

( R i s a c o n s t a n t . ) ('R' is~a~constant.)

By doing this, integration becomes simpler, and comes out to be ( R 2 x 2 ) (R^2 - x^2)

Nishant Rai - 6 years, 1 month ago

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Oh thank u very much, can u tell me the source of the problem and plz the a reply posted by me on ur note.

Kyle Finch - 6 years, 1 month ago

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@Kyle Finch This question came in our mock test for JEE Adanced.

Nishant Rai - 6 years, 1 month ago

The geometric of a hemisphere is 3 R / 8 3R/8 above its center; wouldn't it make sense for the force to be ( Q / 2 ) ( Q / 2 ) 4 π ϵ ( 3 R / 4 ) 2 = Q 2 9 π ϵ R \displaystyle \frac{(Q/2)(Q/2)}{4\pi \epsilon (3R/4)^{2}} = \frac{Q^{2}}{9 \pi \epsilon R} ?

Jake Lai - 6 years, 1 month ago

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