Two fixed, identical conducting plates (α &β ) , each of surface area S are charged to –Q and q, respectively, where Q > q > 0. A third identical plate (γ ), free to move is located on the other side of the plate with charge q at a distance d . The third plate is released and collides with the plate β . Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst β &γ . (a) Find the electric field acting on the plate γ before collision. (b) Find the charges on β and γ after the collision.

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alok kumar singh | Contributor-Level 10

4 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- net electric field at plate γ due to two other plate

From plate 1 ,E₁= $\frac{- Q}{S (2 ε_{0})} t o t h e l e f t$

From plate 2 ,E₂= $\frac{q}{S (2 ε_{0})} t o t h e r i g h t$

Total electric field E= E₁+ E₂ = $\frac{q - Q}{S (2 ε_{0})}$ to the left , if Q>q

electric field at o due to plate α = $\frac{- Q}{S (2 ε_{0})} t o t h e l e f t$

electric field at o due to plate β = $\frac{q 1}{S (2 ε_{0})} t o t h e r i g h t$

electric field at o due to plate γ = $\frac{q 2}{S (2 ε_{0})} t o t h e l e f t$

as the electric field at o is zero therefore

As there is no loss of charge on collision

Q+q= $\frac{Q + q_{2}}{S (2 ε_{0})} = \frac{q_{1}}{S (2 ε_{0})}$

On solving these

q₁= (Q+q/2)= charge on plate β

q₂= (q/2)= charge on plate γ

This is a Long Answer Type Questions as classified in NCERT ExemplarExplanation- net electric field at plate γ due to two other plateFrom plate 1 ,E1= <math> <mfrac> <mrow> <mrow> <mo>-</mo> <mi>Q</mi> </mrow> </mrow> <mrow> <mrow> <mi>S</mi> <mfenced separators="|"> <mrow> <mrow> <mn>2</mn> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> </mfrac> <mi> </mi> <mi>t</mi> <mi>o</mi> <mi> </mi> <mi>t</mi> <mi>h</mi> <mi>e</mi> <mi> </mi> <mi>l</mi> <mi>e</mi> <mi>f</mi> <mi>t</mi> </math> From plate 2 ,E2= <math> <mfrac> <mrow> <mrow> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <mi>S</mi> <mfenced separators="|"> <mrow> <mrow> <mn>2</mn> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> </mfrac> <mi> </mi> <mi>t</mi> <mi>o</mi> <mi> </mi> <mi>t</mi> <mi>h</mi> <mi>e</mi> <mi> </mi> <mi>r</mi> <mi>i</mi> <mi>g</mi> <mi>h</mi> <mi>t</mi> </math> Total electric field E= E1 + E2 = <math> <mfrac> <mrow> <mrow> <mi>q</mi> <mo>-</mo> <mi>Q</mi> </mrow> </mrow> <mrow> <mrow> <mi>S</mi> <mfenced separators="|"> <mrow> <mrow> <mn>2</mn> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> </mfrac> <mi> </mi> </math> to the left , if Q>qelectric field at o due to plate α = <math> <mfrac> <mrow> <mrow> <mo>-</mo> <mi>Q</mi> </mrow> </mrow> <mrow> <mrow> <mi>S</mi> <mfenced separators="|"> <mrow> <mrow> <mn>2</mn> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> </mfrac> <mi> </mi> <mi>t</mi> <mi>o</mi> <mi> </mi> <mi>t</mi> <mi>h</mi> <mi>e</mi> <mi> </mi> <mi>l</mi> <mi>e</mi> <mi>f</mi> <mi>t</mi> </math> electric field at o due to plate β = <math> <mfrac> <mrow> <mrow> <mi>q</mi> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>S</mi> <mfenced separators="|"> <mrow> <mrow> <mn>2</mn> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> </mfrac> <mi> </mi> <mi>t</mi> <mi>o</mi> <mi> </mi> <mi>t</mi> <mi>h</mi> <mi>e</mi> <mi> </mi> <mi>r</mi> <mi>i</mi> <mi>g</mi> <mi>h</mi> <mi>t</mi> </math> electric field at o due to plate γ = <math> <mfrac> <mrow> <mrow> <mi>q</mi> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mi>S</mi> <mfenced separators="|"> <mrow> <mrow> <mn>2</mn> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> </mfrac> <mi> </mi> <mi>t</mi> <mi>o</mi> <mi> </mi> <mi>t</mi> <mi>h</mi> <mi>e</mi> <mi> </mi> <mi>l</mi> <mi>e</mi> <mi>f</mi> <mi>t</mi> </math> as the electric field at o is zero thereforeAs there is no loss of charge on collisionQ+q= <math> <mfrac> <mrow> <mrow> <mi>Q</mi> <mo>+</mo> <msub> <mrow> <mrow> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>S</mi> <mfenced separators="|"> <mrow> <mrow> <mn>2</mn> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>S</mi> <mfenced separators="|"> <mrow> <mrow> <mn>2</mn> <msub> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> </mfrac> </math> On solving these q1= (Q+q/2)= charge on plate βq2= (q/2)= charge on plate γ

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