Q19.2-20PE
Question
Two parallel conducting plates are separated by \(10.0{\rm{ }}cm\), and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential \(8.00{\rm{ }}cm\) from the zero-volt plate (and \(2.00{\rm{ }}cm\) from the other) is \(450{\rm{ }}V\)? (b) What is the voltage between the plates?
Step-by-Step Solution
Verified- Value of the electric field strength is \(E = 5625\;V/m\).
- Value of the voltage between the plates is \(\Delta V = 563\;V\).
The potential difference between two points separated by a distance \(d\) in a homogeneous electric field of magnitude \(E\) is \(\Delta V = Ed{\rm{ }}......(1)\).
Now, calculation for the distance between the plates:
\(\begin{array}{c}d = (10\;cm)\left( {\frac{{1\;m}}{{100\;cm}}} \right)\\d = 0.100{\rm{ }}m\end{array}\)
The plate with the lower potential is taken to be at zero volts.
The electric potential distance \(d = 8.0{\rm{ }}cm\)from the plate with the lower potential is given as
\(V = 450{\rm{ }}Volts\)
(a)
The electric field strength between the two plates is found by solving equation for \(E\):
\(E = \frac{{\Delta V}}{d}\)
Where \(V = 450{\rm{ }}Volts\) is the electric potential a distance \(d = 0.0800\;m\) from the plate with the zero potential is
\(\begin{array}{c}E = \frac{{450\;V}}{{0.0800\;m}}\\E = 5625\;V/m\end{array}\)
Therefore, electric field value is obtained as \(E = 5625\;V/m\).
b)
Use the equation\(\Delta V = Ed\).
Where \(E = 5625\;V/m\) is the electric field strength between the two plates and \(d = 0.0100\;m\) is the distance between the two plates:
\(\begin{array}{c}\Delta V = (5625\;V/m)(0.100\;m)\\\Delta V = 563\;V\end{array}\)
Hence, voltage value is obtained as \(\Delta V = 563\;V\).