Question

One point has a voltage of 9 volts and a second point 5 millimeters away has a voltage of 3 volts What is the strength of the electric field between these points?

Answer 1

Answer:

1,200V/m

Explanation:

I will call the first point A and the second point B, we have:

voltage in A: $V_{A}=9V$

Voltage in B: $V_{B}=3V$

The distance between the points: $d= 5mm = 5x10^{-3}m$

And we calculate the electric field due to the difference of potential as follows:

$E=\frac{V_{A}-V_{B}}{d}$

Substituting known values:

$E=\frac{9V-3V}{5x10^{-3}m}=\frac{6V}{5x10^{-3}m} = 1,200V/m$

the strength of the electric field between these points 1,200V/m