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Question: Five similar condenser plates, each of area A. are placed at equal distance d apart and are connecte...

Five similar condenser plates, each of area A. are placed at equal distance d apart and are connected to a source of e.m.fE as shown in the following diagram. The charge on the plates 1 and 4 will be

A

ε0Ad,−2ε0Ad\frac{\varepsilon_{0}A}{d},\frac{- 2\varepsilon_{0}A}{d}

B

ε0AVd,−2ε0AVd\frac{\varepsilon_{0}AV}{d},\frac{- 2\varepsilon_{0}AV}{d}

C

ε0AVd,−3ε0AVd\frac{\varepsilon_{0}AV}{d},\frac{- 3\varepsilon_{0}AV}{d}

D

ε0AVd,−4ε0AVd\frac{\varepsilon_{0}AV}{d},\frac{- 4\varepsilon_{0}AV}{d}

Answer

ε0AVd,−2ε0AVd\frac{\varepsilon_{0}AV}{d},\frac{- 2\varepsilon_{0}AV}{d}

Explanation

Solution

Here five plates are given, even number of plates are

connected together while odd number of plates are connected together so, four capacitors are formed and they are in parallel combination, hence redrawing the figure as shown below.

Capacitance of each

Capacitor is C=ε0AdC = \frac{\varepsilon_{0}A}{d}

Potential difference across each capacitor is V

So charge on each capacitor Q=ε0AdVQ = \frac{\varepsilon_{0}A}{d}V

Charge on plate (1) is +ε0AVd+ \frac{\varepsilon_{0}AV}{d}

While charge on plate 4 is −ε0AVd×2- \frac{\varepsilon_{0}AV}{d} \times 2 =−2ε0AVd.= - \frac{2\varepsilon_{0}AV}{d}.