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anastassius [24]
4 years ago
13

A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 μC on each plate. The plates have a

separation of 0.313 mm. What is the potential difference between the plates?
Physics
1 answer:
butalik [34]4 years ago
7 0

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance C=260\ pF

Charge q=0.155\ \mu\ C

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

V= \dfrac{Q}{C}

Where, Q = charge

C = capacitance

Put the value into the formula

V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}

V=596.2\ volts

Hence,The potential difference between the plates is 596.2 volts.

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