Question

Two parallel conducting plates are separated by 10.0 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 cm from the zero volt plate

Answer 1

Complete Question

Two parallel conducting plates are separated by 10.0 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 cm from the zero volt plate(and 2.00 cm from the other) is 450 V?

Answer:

$V'=562.5v$

Explanation:

From the question we are told that:

Separation distance $d=10cm$

Voltage at 8cm $V_8=450v$

Generally the equation for Voltage is mathematically given by

$|V|=|E.d|$

Where

E=electric field

Therefore

At $d=0.8$

 $(450-0)V=E*(0.08m)$

 $E=\frac{450}{0.08}$

 $E=5625v/m$

Therefore

 At $d=10$

 $V'=Ed$

 $V'=5625*0.1m$

 $V'=562.5v$