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In the above question if the charge distributed is changed as shown in figure, the force on q is

Asked by anshuman.anshuman090 | 06 Apr, 2019, 09:12: PM

Expert Answer

Let us consider a small element dl in the ring that gives a repulsive force F to the charge q placed at the centre.

Force F can be resolved along horizontal direction as Fsinθ and in vertical direction as Fcosθ.

Horizontal component Fsinθ will be canceld out if we consider similar element of the ring in right side.

Hence the net force F₊ due to positive charge distribution is given by

begin mathsize 12px style F subscript plus space equals space integral subscript negative straight pi divided by 2 end subscript superscript plus straight pi divided by 2 end superscript F cos theta space space equals space integral subscript negative straight pi divided by 2 end subscript superscript plus straight pi divided by 2 end superscript fraction numerator q space lambda subscript 2 d l over denominator 4 πε subscript straight o straight r squared end fraction cos theta space equals space fraction numerator q lambda subscript 2 over denominator 4 πε subscript straight o straight r squared end fraction integral subscript negative straight pi divided by 2 end subscript superscript plus straight pi divided by 2 end superscript cos theta space r d theta space equals space fraction numerator q lambda subscript 2 over denominator 4 πε subscript straight o straight r end fraction integral subscript negative straight pi divided by 2 end subscript superscript plus straight pi divided by 2 end superscript cos theta space d theta space equals space fraction numerator 2 q lambda subscript 2 over denominator 4 πε subscript straight o straight r end fraction end style

..........................(1)

In the above eqn.(1), λ₂ is the charge density per unit length.

Similarly attractive force F_- due to negative charge distribution is also proportional to charge density λ₁

Both tthe forces are acting in same direction. Hence total force =

begin mathsize 12px style fraction numerator 2 q over denominator 4 πε subscript straight o straight r end fraction open parentheses lambda subscript 1 space plus space lambda subscript 2 close parentheses space N e w t o n end style

Answered by Thiyagarajan K | 06 Apr, 2019, 10:57: PM

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