Question

What is the electric field between the plates of a capacitor that has a charge of 14.85 microC and voltage difference between the plates of 59.98 Volts if the plates are separated by 2.17 mm?

Answer 1

The electric field between two parallel plates is given by the equation:

`E=(V)/(d)`

where V is the difference of potential and d is the distance between the plates. In this case the difference of potential is 59.98 V and the distance between the plates is 2.17 mm, then we have:

`\begin{gathered} E=(59.98)/(2.17*10^(-3)) \\ E=2.76*10^4 \end{gathered}`

Therefore, the electric field is:

`E=2.76*10^4\text{ }(N)/(C)`