A parallel-plate capacitor is formed by two plates, each of area 100 cm2 separated by a distance of 1 mm. A dielectric of dielectric constant 5.0 and dielectric strength 1.9×107Vm−1 is filled between the plates. Find the maximum charge that can be stored on the capacitor without causing any dielectric breakdown.

A
$8 .4 \times 10^{- 6} C$
B
$9 \times 10^{- 6} C$
C
$11 \times 10^{- 6} C$
D
10 C

Solution:

If Q be the charge on the capacitor, the surface charge density is $σ = Q / A$

and the electric field is $\frac{Q}{{KAε}_{0}}$ .

This should not exceed the dielectric strength $1 .9 \times 10^{7} {Vm}^{- 1}$ .

Thus, the maximum charge is given by $\frac{Q}{{KAε}_{0}} = 1 .9 \times 10^{7} {Vm}^{- 1}$
Or, $Q = {KAε}_{0} \times 1 .9 \times 10^{- 7} {Vm}^{- 1}$
$= (5.0) (10^{- 2} m^{2}) (8.85 \times 10^{- 12} {Fm}^{- 1}) \times (1.9 \times 10^{7} {Vm}^{- 1}) = 8.4 \times 10^{- 6} C$
.

Solution:

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