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NCERT Exemplar · Q28

Q.Two fixed, identical conducting plates α\alpha and β\beta, each of surface area SS, are held parallel to each other. Plate α\alpha (on the left) carries charge −Q-Q and plate β\beta (in the middle) carries charge qq, with Q>q>0Q > q > 0. A third identical plate γ\gamma, free to move, is placed parallel to them on the far side of β\beta (to the right), at a distance dd from β\beta; initially γ\gamma is uncharged. The plate γ\gamma is released and slides toward β\beta, striking it. The collision is elastic, and the contact time is long enough for charge to redistribute between β\beta and γ\gamma while they touch.

(a) Find the electric field acting on plate γ\gamma just before the collision.
(b) Find the charges on β\beta and γ\gamma after the collision.
(c) Find the speed of plate γ\gamma after the collision, once it has travelled a distance dd back from plate β\beta.
Rajasthan RbseLong· 5mImportance★★★★★
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Each charged plate of charge σS\sigma S produces a uniform field charge2ε0S\dfrac{\text{charge}}{2\varepsilon_0 S} on each side. Superposing the fields of α\alpha and β\beta gives the field on γ\gamma before impact. During contact β\beta and γ\gamma merge into one conductor, so the charge redistributes onto the outer faces, giving qβ=Q+q2q_\beta=\frac{Q+q}{2} and qγ=q−Q2q_\gamma=\frac{q-Q}{2}. After the collision the (uniform, hence constant) net force on γ\gamma does work F dF\,d, which equals its kinetic energy, fixing the speed.

Model

A large plate carrying charge ± ⁣  \pm\!\;(magnitude) over area SS acts like a charged sheet producing a uniform field of magnitude ∣charge∣2ε0S\dfrac{|\text{charge}|}{2\varepsilon_0 S} on each side, directed away from the plate if positive, toward it if negative. Take +x+x to the right (from α\alpha toward γ\gamma).

(a) Field on γ\gamma before collision (γ\gamma uncharged)

At γ\gamma's location (to the right of both α\alpha and β\beta):

  • Due to α (−Q)\alpha\,(-Q): magnitude Q2ε0S\dfrac{Q}{2\varepsilon_0 S}, pointing toward α\alpha (left).
  • Due to β (+q)\beta\,(+q): magnitude q2ε0S\dfrac{q}{2\varepsilon_0 S}, pointing away from β\beta (right).

Net (taking left as the resultant direction since Q>qQ>q):

Eγ=Q2ε0S−q2ε0S=Q−q2ε0Sdirected toward the plates (left).E_\gamma = \frac{Q}{2\varepsilon_0 S} - \frac{q}{2\varepsilon_0 S} = \frac{Q-q}{2\varepsilon_0 S}\quad\text{directed toward the plates (left).}

(b) Charges after the collision

While β\beta and γ\gamma touch they form a single conductor of total charge q+0=qq + 0 = q, sitting next to plate α\alpha (charge −Q-Q). For two separated conductors, the two outer faces carry equal charge (total of both)2=−Q+q2\dfrac{(\text{total of both})}{2}=\dfrac{-Q+q}{2}, and the facing surfaces carry ±Q+q2\pm\dfrac{Q+q}{2}. Since the merged β\beta–γ\gamma conductor has no charge on its internal contact interface:

  • β\beta (inner plate, facing α\alpha) carries the facing-surface charge:   qβ=Q+q2.\;q_\beta = \dfrac{Q+q}{2}.
  • γ\gamma (outer plate) carries the outer-surface charge:   qγ=q−Q2  (<0).\;q_\gamma = \dfrac{q-Q}{2}\;(<0).

Check: qβ+qγ=Q+q2+q−Q2=q.q_\beta + q_\gamma = \dfrac{Q+q}{2}+\dfrac{q-Q}{2} = q. ✓

(c) Speed of γ\gamma after travelling dd

After the collision γ\gamma carries qγ=q−Q2q_\gamma=\dfrac{q-Q}{2}. The field on γ\gamma from the other two plates:

  • Due to α (−Q)\alpha\,(-Q): Q2ε0S\dfrac{Q}{2\varepsilon_0 S} toward α\alpha (left).
  • Due to β (Q+q2)\beta\,\big(\tfrac{Q+q}{2}\big): Q+q4ε0S\dfrac{Q+q}{4\varepsilon_0 S} away from β\beta (right). …

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