Two identical conducting plates A and B with charges -Q (- 2 C ) and + q (1C) respectively are located parallel to each other at a small distance. Another identical plate C with mass m and charge +Q (

Question

Two identical conducting plates A and B with charges -Q (- 2 C ) and + q (1C) respectively are located parallel to each other at a small distance. Another identical plate C with mass m and charge +Q ( $\frac{3}{2} C$ ) is situated parallel to the original plates at a distance d from plate B. (see figure). Area of each plates is S. The plate C is released and can move freely, collides elastically with plate B. [Neglect gravity and boundary effects]. Assume that the charge has enough time to redistribute. The velocity v of plate C after the collision at a distance d from the plate B is $K \frac{d}{m ε _{0} S}$ . Find value of K.[ plates A and B remain fixed ]

TestHub · Accepted Answer

At any instant, the force on plate C is $F = \frac{( Q - q ) Q}{2 S ε _{0}}$ .

The work done is $W = \frac{( Q - q ) Q d}{2 S ε _{0}}$ .

After the collision, the charge on plates A, B, and C becomes:

Work done is $W = \frac{( q /2 ) ^{2} d}{2 S ε _{0}}$ .

Total work done:

$W_{t o t a l} = \frac{( Q - q ) Q d}{2 S ε _{0}} + \frac{( \frac{q}{2} ) ^{2} d}{2 S ε _{0}} = \frac{d}{2 S ε _{0}} [Q^{2} - qQ + (\frac{q}{2})^{2}] = \frac{d}{2 S ε _{0}} [(Q - \frac{q}{2})^{2}]$

Total work done = Change in KE

$\frac{1}{2} m v^{2} = \frac{d}{2 S ε _{0}} (Q - \frac{q}{2})^{2}$

$\Rightarrow v = \frac{d}{m S ε _{0}} (Q - \frac{q}{2})$

$\Rightarrow v = \frac{d}{m S ε _{0}} [\frac{3}{2} - \frac{1}{2}] = 1$

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