Question

Two 4.634 cm by 4.634 cm plates that form a parallel-plate capacitor are charged to +/- 0.947 nC.What is the electric field strength inside the capacitor if the spacing between the plates is 1.624 mm?

Answer 1

The electric field between two parallel plate is given by:

`E=4\pi k(Q)/(A)`

where k is the Coulomb's constant, Q is the charge on the plates and A is the area of each plate. In this case we have that:

`\begin{gathered} Q=0.947*10^(-9) \\ A=(4.634*10^(-2))^2 \end{gathered}`

Then we have:

`\begin{gathered} E=4\pi(9*10^9)((0.947*10^(-9)))/((4.634*10^(-2))^2) \\ E=4.99*10^4 \end{gathered}`

Therefore, the electric field is:

`E=4.99*10^4\text{ }(N)/(C)`