Question

a parallel plate capacitor has circular plates of radius 8cm and plate separation 1 mm. what will be the charge on the plates if a potential difference of 100 V is applied? ​

Answer 1

Given:

Radius of circular parallel plate capacitor (r) = 8 cm = 0.08 m

Seperation between plate (d) = 1 mm = 0.001 m

Potential difference applied (V) = 100 V

To Find:

Charge on the plates (Q)

Answer:

Capacitance of parallel plate capacitor:

$\boxed{ \boxed{ \bf{C = \dfrac{Q}{V} = \dfrac{\epsilon_0 A}{d}}}}$

A → Area of parallel plate

As parallel plates are circular. So, area of parallel plates of capacitor is given as πr²

Thus,

$\rm \leadsto \dfrac{Q}{V} = \dfrac{\epsilon_0 \pi { r}^{2} }{d} \\ \\ \rm \leadsto Q = \dfrac{\epsilon_0 \pi { r}^{2} }{d} \times V \\ \\ \rm \leadsto Q = \dfrac{8.85 \times {10}^{ - 12} \times 3.14 \times {(8 \times {10}^{ - 2}) }^{2} }{1 \times {10}^{ - 3} } \times 100 \\ \\ \rm \leadsto Q = \frac{8.85 \times {10}^{ - 12} \times 3.14 \times 64 \times {10}^{ - 4} \times {10}^{2} }{ {10}^{ - 3} } \\ \\ \rm \leadsto Q = 1778.496 \times {10}^{ - 12 - 4 + 2 + 3 } \\ \\ \rm \leadsto Q = 1778.496 \times {10}^{ - 11 } \\ \\ \rm \leadsto Q = 17.78 \times {10}^{ - 9} \: C \\ \\ \rm \leadsto Q = 17.78\: n C$

∴ $\boxed{\mathfrak{Charge \ on \ the \ plates \ (Q) = 17.78 \ nC}}$

Answer 2

C = εA/d

= επr²/d

= 8.85 × 10^-12 × 3.14 × (0.08)^2/0.001

= 1.78 × 10^-10 F

Q = CV

= 1.78 × 10^-10 × 100

= 1.78 × 10^-8 C